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AUG 1 



2 9 



1950 




Southern Branch 
of the 

University of California 



Los Angeles 



Form L 1 



This book is DUE on the last date stamped below 

JUL 8 1*2? 



JUL 1 8 W 



I 2 



2 



7 1 
8 




'JUL 2 9 1953 
MAN 2 7 1954 



1962 
- 2 3 1979 



n L-9-15m-8,'26 



GENERAL MATHEMATICS 



BY 

RALEIGH SCHORLING 

IN CHARGE OF MATHEMATICS, THE LINCOLN SCHOOL OF TEACHERS 
COLLEGE, NEW YORK CITY 



AND 



WILLIAM DAVID REEVE 

TEACHERS' TRAINING COURSE IN MATHEMATICS IN THE COLLEGE 
OF EDUCATION, ANI> HEAD OF THE MATHEMATICS DEPART- 
MENT IN THE UNIVERSITY HIGH SCHOOL 
THE UNIVERSITY OF MINNESOTA 



GINN AND COMPANY 

BOSTON NEW YORK CHICAGO LONDON 
ATLANTA DALLAS COLUMBUS SAN FRANCISCO 



2883 



171-50 

m 



COPYRIGHT, 1919, BY 
RALEIGH SCHORLING AND WILLIAM DAVID REEVE 



ALL, RIGHTS RESERVED 
819.11 



7171 



gtbenaum 



GINN AND COMPANY PRO- 
PRIETORS BOSTON U.S.A. 



''A- 



PREFACE 

The purpose of this book, as implied in the introduction, 
is as follows : to obtain a vital, modern scholarly course in 
introductory mathematics that may serve to give such careful 
training in quantitative thinking and expression as well- 
informed citizens of a democracy should possess. It is, of 
course, not asserted that this ideal has been attained. Our 
achievements are not the measure of our desires to im- 
prove the situation. There is still a very large " safety 
factor of deud wood" in this text. The material purposes 
to present such simple and significant principles of algebra, 
geometry, trigonometry, practical drawing, and statistics, 
along with a few elementary notions of other mathematical 
subjects, the whole involving numerous and rigorous appli- 
cations of arithmetic, as the average man (more accurately 
the modal man) is likely to remember and to use. There 
is here an attempt to teach pupils things worth knowing 
and to discipline them rigorously in things worth doing. 

The argument for a thorough reorganization need not 
be stated here in great detail. But it will be helpful to 
enumerate some of the major errors of secondary-mathe- 
matics instruction in current practice and to indicate 
briefly how this work attempts to improve the situation. 
The following serve to illustrate its purpose and program : 
1. The conventional first-year algebra course is charac- 
terized by excessive formalism; and there is much drill 
work largely on nonessentials. The excessive formalism is 



iv GENERAL MATHEMATICS 

greatly reduced in this text and the emphasis placed on 
those topics concerning which there is general agreement, 
namely, function, equation, graph, and formula. The time 
thus gained permits more ample illustrations and applica- 
tions of principles and the introduction of more significant 
material. 

2. Instead of crowding the many difficulties of the 
traditional geometry course into one year, geometry in- 
struction is spread over the years that precede the formal 
course, and the relations are taught inductively by experi- 
ment and by measurement. Many foreign schools and an 
increasing number of American schools proceed on this 
common-sense basis. This gives the pupil the vocabulary, 
the symbolism, and the fundamental ideas of geometry. 
If the pupil leaves school or drops mathematics, he never- 
theless has an effective organization of geometric relations. 
On the other hand, if he later pursues a formal geometry 
course, he can work far more effectively because he can 
concentrate on the logical organization of space relations 
and the formal expression of these relations. The longer 
" time exposure " minimizes the difficulties met in begin- 
ning the traditional geometry courses and avoids the 
serious mistake of forcing deductive logic and philosophic 
criticism in these early years. 

3. The traditional courses delay the consideration of 
much interesting and valuable material that the field of 
secondary mathematics has to offer, and which may well 
be used to give the pupil very early an idea of what 
mathematics means and something of the wonderful scope 
of its application. The material of the seventh, eighth, and 
ninth years is often indefensibly meaningless when com- 
pared with that of many foreign curricula. Trigonometry, 



containing many easy real problems, furnishes a good ex- 
ample of this delay. Other examples are found in the 
use of logarithms, the slide rule, standardized graphical 
methods, the notion of function, the common construction 
of practical drawing, the motivation of precise measure- 
ment, a study of the importance of measurement in modern 
life, and the introductory ideas of the calculus. It appears 
that the mathematics student should be given an oppor- 
tunity to use these important tools very early in his study. 
They lend to the subject a power and interest that drills on 
formal material cannot possibly give. 

Particular emphasis is given to graphical representation 
of statistics. The growing complexity of our social life 
makes it necessary that the intelligent general reader 
possess elementary notions of statistical methods. The 
hundreds of articles in the current magazines so exten- 
sively read demand an elementary knowledge of these 
things in order that the pupil may not remain ignorant of 
the common, everyday things of life. Brief chapters on 
logarithms and the slide rule have been introduced in 
order that a greater number of students may use these 
practical labor-saving devices and in order that these devices 
may function in the student's subsequent work, whether 
in everyday life or in the classroom. Actual classroom 
experience with these chapters has proved them to be 
relatively simple and good material for eighth-grade and 
ninth-grade students. 

4. Mathematics needs to be reorganized on the side 
of method. The information we now possess of individual 
differences and effective devices in supervised study should 
make the study of mathematics more nearly a laboratory 
course, in which more effective work can be done. 



vi ( i KN KJ{ A L M ATI 1 KM AT1CS 

5. The teaching of algebra, geometry, and trigonometry 
in separate fields is an artificial arrangement that does not 
permit the easy solution of problems concerning projects 
that correlate with problems met in the physical and bio- 
logical sciences or the manual and fine arts. To reject the 
formalism of algebra, to delay the demands of a logical 
unit in geometry, and to present the simple principles of 
the various branches of mathematics in the introductory 
course opens the door to a greater variety of problems that 
seem to be real applications. The pupil sees the usefulness 
of the various modes of treatment of the facts of quantity. 
Power is gained because the pupil is equipped with more 
tools, in that the method of attack is not limited to one field. 

6. One of the most curious characteristics of American 
secondary-mathematics instruction is the obscurity in the 
teaching of the function notion. It is generally agreed 
that functional thinking (the dependence of one magnitude 
upon another) constitutes one of the most fundamental 
notions of mathematics. Because of the interrelations of 
the equation, the formula, the function, the graph, and the 
geometric relations inductively acquired, the material is 
easily correlated around the function idea as the organizing 
and unifying principle. The function concept (implicitly or 
explicitly) dominant throughout helps to lend concreteness 
and coherence to the subject. However, it would be false 
to assume that this material is presented to establish the 
principle of correlation. On the contrary, it happens that 
correlation around the function notion, though incidental, 
is a valuable instrument for accomplishing the larger aim, 
which is to obtain a composite introductory course in mathe- 
matics that all future citizens of our democracy should be 
required to take as a matter of general scholarship. 



PREFACE vii 

7. The traditional reticence of texts has made mathe- 
matics unnecessarily difficult for pupils in the early years. 
The style of this book, though less rigidly mathematical, is 
more nearly adapted to the pupils' mental age. The result 
is a misleading length of the book. The book can easily be 
taught in a school year of approximately one hundred and 
sixty recitations. In the typical high school it will be taught 
in the first year. (The Minnesota high schools taught it 
in this grade, five recitations per week.) In schools which 
control the seventh and eighth years the following are 
also possibilities which have been tested by the authors 
and cooperating teachers : (1) in the eighth year with daily 
recitations ; (2) half of the book in the eighth year and the 
remainder in the ninth year, with three recitations per week 
(it was so used in the Lincoln School) ; (3) the course 
may be started in the seventh year provided the class has 
achieved good results in previous arithmetic work. 

Specific references are given where material which is 
not of the common stock has been taken consciously, the 
purpose, however, being chiefly to stimulate pupils and 
teachers to become familiar with these books for reasons 
other than the obligations involved. Something of human 
interest is added by relating some of the well-known stories 
of great mathematicians. We are indebted to Professor 
David Eugene Smith on questions relating to historical 
material. In our thinking we are particularly indebted to 
Professors Nunn, Smith, Breslich, and Myers. We shall 
be obliged to all teachers who may think it worth while 

to point out such errors as still exist. 

THE AUTHORS 



CONTENTS 

CHAPTER PAGE 

I. THE EQUATION 1 

Solving an equation 6 

Translation of an equation . . 12 

Solution of verbal problems 16 

Axioms 21 

II. LINEAR MEASUREMENT. THE EQUATION APPLIED TO 

LENGTH 26 

Different units of length 28 

Squared paper 32 

Sum of two segments ; geometric addition 36 

Polygons * 44 

III. PROPERTIES OF ANGLES 47 

Notation for reading angles 50 

Measurement of angles ; the protractor 54 

Measuring angles ; drawing angles 56 

Comparison of angles - 59 

Geometric addition and subtraction of angles 61 

Parallel lines 68 

How to construct a parallelogram 70 

IV. THE EQUATION APPLIED TO AREA 74 

Formula 78 

Formula for the area of a parallelogram 79 

Geometric interpretation of products 85 

Algebraic multiplication 89 

The accuracy of the result 93 

V. THE EQUATION APPLIED TO VOLUME 98 

Measurement of volume 99 

Formula for the volume of a rectangular parallelepiped . 99 

ix 



x GE^'EKAL MATHEMATICS 

CHAPTER PAGE 

Formula for the volume of a cube 102 

Exponents 102 

Application of algebraic principles to geometric figures . 105 

VJ. THE EQUATION APPLIED TO FUNDAMENTAL ANGLE 

RELATIONS Ill 

The sum of all the angles about a point on one side 

of a straight line 112 

Sum of all the angles about a point in a plane . . . 113 

Supplementary angles ; supplement 116 

Complementary angles 119 

Vertical angles 122 

Important theorems relating to parallel lines .... 126 

VII. THE EQUATION APPLIED TO THE TRIANGLE .... 130 

The sum of the interior angles 131 

Right triangle -. . 135 

Exterior angles of a triangle 139 

The construction of triangles 142 

VIII. POSITIVE AND NEGATIVE NUMBERS. ADDITION AND 

SUBTRACTION , 150 

Use of signs 151 

Geometric representation of positive numbers. Origin . 152 

Geometric representation of negative numbers . . . 153 

Algebraic addition 162 

Subtraction illustrated by the number scale .... 170 

Algebraic subtraction 171 

Subtraction of polynomials 173 

IX. POSITIVE AND NEGATIVE NUMBERS. MULTIPLICATION 

AND DIVISION. FACTORING 178 

Geometric illustration of law of signs 178 

Law of signs illustrated by a balanced bar 180 

Multiplication of positive and negative numbers . . . 182 

Special products 192 

Law of signs in division 195 

Factoring 198 



CONTENTS xi 

CHAPTER PAGE 

Distinction between identity and equation 204 

Use of factoring in identities for calculating areas . . 205 

X. GRAPHICAL REPRESENTATION OF STATISTICS; THE 

GRAPH OF A LINEAR EQUATION 214 

Pictograms 214 

Practice in interpreting the bar diagram 222 

How to construct a bar diagram 224 ' 

Practice in interpreting graphic curves 231 

How the. graphic curve is drawn 233 

Normal distribution 257 

Symmetry of a curve 259 

Graph of constant cost relations 262 

Graphs of linear equations 263 

XI. GAINING CONTROL OF THE FORMULA; GRAPHICAL 

INTERPRETATION OF FORMULAS 273 

Solving a formula 276 

Graphical illustration of a motion problem 283 

Translating rules of procedure into formulas .... 287 

Graph of the centigrade-Fahrenheit formula .... 288 

Evaluating a formula 290 

XII. FUNCTION 299 

Graph of a function 301 

Solving the function set equal to zero 304 

Direct variation 305 

Graphing direct variation 308 

Graphing inverse variation . . . ... . . . . . 311 

XIII. SIMILARITY; CONSTRUCTION OF SIMILAR TRIANGLES . 314 

Summary of constructions for similar triangles . . . 317 

Algebraic problems on similar figures ...... 319 

Proportion 322 

Construction of a mean proportional 332 

Fourth proportional construction 334 

Verbal problems solved by proportion 336 

Proportionality of areas 341 



xii GENERAL MATHEMATICS 

CHAPTER PAGE 

XIV. INDIRECT MEASUREMENT; SCALE DRAWINGS; TRIG- 
ONOMETRY 345 

Similar right triangles :;.",:, 

Trigonometric ratios :;f>!t 

Table of trigonometric ratios 361 

Verbal trigonometry problems 362 

XV. THEORY AND APPLICATION OF SIMULTANEOUS 

LINEAR EQUATIONS 367 

Graphic solution 369 

Algebraic methods for solving simultaneous linear 

equations 373 

Summary of methods of elimination 379 

Classified verbal problems 384 

XVI. GEOMETRIC AND ALGEBRAIC INTERPRETATION OF 

ROOTS AND POWERS 390 

The theorem of Pythagoras 397 

Constructing the square root of a number .... 404 
Fractional exponents another means of indicating 

roots and powers 412 

XVII. LOGARITHMS * .... 424 



Logarithms defined , 427 

Exponential equations 443 

Interest problems solved by logarithms 444 

XVIII. THE SLIDE RULE . 449 

Square roots found by means of the slide rule . . . 455 

Verbal problems solved by the slide rule 458 

XIX. QUADRATIC FUNCTIONS; QUADRATIC EQUATIONS . 462 

How to solve a quadratic equation graphically . . . 465 

The parabola 467 

More powerful methods of solving quadratic equations 471 

Maxima and minima algebraically determined . . . 479 

INDEX 481 



INTRODUCTION 

The movement to provide an introductory course in general 
mathematics is a part of an extensive movement toward mak- 
ing the materials of study in secondary education more concrete 
and serviceable. The trend of education expresses a determi- 
nation that the seventh, eighth, and ninth school years should 
be enriched by the introduction of such significant experiences 
of science, civics, art, and other knowledge of human life as 
all enlightened citizens of a democracy should possess. The 
work of these grades cannot be liberalized by "shoving down" 
the conventional material a year or so. The reorganization 
must be more fundamental in order to revitalize and socialize 
the mathematics of these grades. 

Competent authorities in mathematics have from time to time 
asserted, first, that American secondary-mathematics teaching 
has been characterized by a futile attempt to induce all pupils 
to become technical college mathematicians. Secondly, that 
instead of giving pupils an idea of the real meaning of mathe- 
matics and the wide range of its applications, they are forced 
to waste a great deal of time on abstract work in difficult 
problems in radicals, fractions, factoring, quadratics, and the 
like, which do not lead to anything important in mathematics. 
And, thirdly, that this meaningless juggling of symbolism fails 
to meet the needs of the great number of pupils who go rather 
early into their careers ; it also wastes time and effort on the 
part of pupils with especial ability in the subject, who ought 
to get an early insight into the scope and power of the real 
science of mathematics. 



xiv GENERAL MATHEMATICS 

Quantitative thinking and expression play so large a part in 
human experience that proper training in these matters will 
always be important. The growing complexity of social and 
industrial life is responsible for corresponding changes in the 
use of quantitative relationships. Old applications' in many 
instances are disappearing, but new ones growing out of present- 
day relations are being introduced to take their places. These 
changes require a new kind of introductory text in mathematics. 
Action is forced by the demand that there shall be justification 
of the time and effort given to each subject and each item in 
the subject. New subjects which appear necessary in the proper- 
training for citizenship are crowding the curriculum. Mathe- 
matics too must justify its place " in the sun " by a thorough 
reorganization that will meet modern needs. This is what is 
meant by revitalizing mathematics. 

The practical administrator will be impressed by the fact 
that this program raises no administrative difficulties. The 
pupil may be expected to develop greater power in algebra, 
because the elimination of material which wastes time and 
effort has made possible the emphasizing of the topics concern- 
ing which there is general agreement. The supplementary 
material which is drawn from the other subjects constitutes a 
preparation for further study in these fields ; for example, the 
text gives the pupil the vocabulary, the symbolism, and many 
of the ideas of plane geometry. 

This type of introductory course should appeal to the pro- 
gressive educator because of a number of other features. The 
" problem method" of teaching is followed throughout. Ration- 
alized drills are provided in abundance. The course has been 
used in mimeographed form by experienced teachers. Scores 
of prospective teachers have found the treatment simple and 
easy to present. Inexperienced teachers have gone out into 
difficult situations and have taught the material with satis- 
faction. Pupils following this course have made better progress 



INTRODUCTION xv 

than pupils following the traditional course, and both, pupils 
and teachers manifest a degree of interest seldom seen in the 
ordinary class in mathematics. The tests prepared by the 
authors will save time for teachers and enable them, if they 
desire, to diagnose their own situations and to compare their 
results with those obtained in other institutions using the same 
material. If the question is raised as to what students com- 
pleting such a course will do when they get to college, it may 
be replied that enough of them have already entered college 
to convince the unbiased that they experience no handicap. 
The more important point, however, is that such a course 
enables one to understand and to deal with the quantitative 
world in which he lives. 

This course in reorganized introductory mathematics, 
although but a part of a large movement in secondary educa- 
tion which looks toward more concrete teaching and more 
serviceable materials of study, has a further highly significant 
aspect. It is a potent and encouraging evidence that high- 
school teachers have become students of their own teaching, 
and as a result are preparing their own textbooks in the midst 
of real teaching situations, as the outcome of intelligent con- 
'structive experimentation. 

Probably very few books have been subjected previous to 
publication to such thorough tests of teaching situations. The 
authors have been shaping this course for many years. During 
the last three years the manuscript as originally accepted by 
the publishers has been taught in mimeograph form to more 
than a thousand pupils distributed in a selection of typical 
schools, among these being the following : Minneapolis Central 
High School (large city high school), Bremer Junior High 
School, Seward Junior High School, University of Minnesota 
High School, Owatonna High School, Mabel High School 
(small town), and the Lincoln School of Teachers' College. 
Numerous consultations with the teachers in these schools 



xvi GENERAL MATHEMATICS 

resulted in many valuable suggestions which contributed directly 
toward making the text easily teachable. 

Each of the authors has taught secondary mathematics for 
more than ten years in large public and private schools. They 
have, supervised many teachers in training ; they have taught 
teacher-training courses, and each during most of this time has 
had unusual opportunities for free experimentation. The text 
may be regarded by fellow teachers of mathematics as a report 
which shows in organization and subject matter the things that 

have seemed most useful. 

LOTUS D. COFFMAN 

OTIS W. CALDWELL 



FIRST YEAR 



CHAPTER I 

THE EQUATION 

1. A problem introducing the use of the equation. In 

order to find the weight of a bag of candy, it was placed 
on one pan of perfectly balanced scales (Fig. 1). The 
candy, together with a 4-ounce weight, balanced 10 oz. 
of weights on the 
other pan. How 
much did the bag 
of candy weigh ? 

It is a familial- 
principle of bal- 
anced scales that 
if the same number 
of ounces be taken 
from each pan, the 
balance is not disturbed. Hence, if we suppose that a 
4-ounce weight could be removed from each pan, the 
candy would be balanced by 6 oz. 

This solves the problem, but let us analyze it a little 
further. The important fact in the situation above is that 
an unknown number of ounces of candy plus 4 oz. in 
one pan balances 10 oz. in the other pan. If we agree 

1 




FIG. 1. THE BALANCED SCALES ILLUSTRATE 
THE MEANING OF AN EQUATION 



2 GENERAL MATHEMATICS 

to let the letter w represent the number of ounces of 
weight in the bag of candy and use the sign of equality 
(=) to denote the perfect balance of the scales, the pre- 
ceding mathematical fact may be conveniently translated 
into the following expression : w + 4 = 10, where w + 4 
denotes the weight in the left pan and 10 the weight in 
the right pan. The abbreviated ("shorthand") statement, 
w + 4=10, expresses equality and is called an equation. 
The number to the left side of the equality sign is called 
the left member of the equation, the number to the right 
is the right member. 

Just as the scales will balance if the same number of 
ounces are taken from each pan, so ive may subtract the 
same number from both aides of an equation and get another 
f'/Hxtion. In the preceding problem the written work may 

be. arranged thus: 

f number of ounces of weight 
Let to = J. . ., 

^ in the bag of candy. 

Then w + 4 = 10 

4=4 
Subtracting 4 from each "I ^ 

member of the equation, J 
Thus, the- bag of candy weighs 6 oz. 

The preceding problem illustrates the principle that if 
the same number be subtracted from both members of an 
equation, the remainders are equal ; that is, another equation 
is obtained. [Subtraction Law] 

EXERCISES 

Find the value of the unknown numbers in the following 
equations, doing all you can orally : 

1. x + 2 = 6. 4. or + 11 = 18. 7. x + 10 = 27. 

2. x + 6 = 10. 5. x + 13 = 23. 8. x + 14 = 21. 

3. x + 7 = l3. 6. z+9 = 26. 9. x -f 33 = 44. 



THE EQUATION 



3 



2. The importance of the equation. The equation is a 
very important tool for solving problems in the mathe- 
matical sciences. The equation gives us a new method of 
attack on a problem, enabling us to solve many problems 
which would be very difficult, if not impossible, without 
its use. 

3. Method of studying the nature of the equation. In 

making a study of the equation we shall continue by con- 
sidering some very simple problems in order that we may 
clearly understand the laws which are involved. If these 
laws are mastered in connection with the simple cases, it 
will be easy to apply the equation as a tool for solving 
the more complicated and difficult problems. In the next 
article we shall continue to interpret the equation by 
considering a problem in weighing. 

4. Division Law. Two equal but unknown weights, to- 
gether with a 1-pound weight, just balance a 16-pound and 
a 1-pound weight to- 
gether (Fig, 2). How 

heavy is each un- 
known weight? 

Let p equal the 
number of pounds in 
one of the unknown 
weights. Suppose that 
1 Ib. be removed from 
each pan, leaving *2p F IG . 2. THE BALANCED SCALES MAT UK USED 
DOUnds in the left T(? ILLUSTRATE THE SUBTRACTION LAW AND 

THE DIVISION LAW i 

pan balancing the re- 
maining 16 Ib. in the right pan. Then, if 2 p pounds 
balances 16 Ib., p pounds (one half of the weight in the 
left pan) must balance 8 Ib. (one half of the weight" in 




4 GENERAL MATHEMATICS 

the right pan). By the use of the equation the discussion 
may be written in the following brief form: 

f This is a translation of the tirst 
" J \ sentence of the problem. 

Subtracting 1 from "1 - 

each member, / ^ 

Dividing each member of the equation by 2, 
p=8. 

This problem illustrates the principle that if both mem- 
bers of an equation are divided by the name number (exclud- 
ing division by zero, to be explained later), the quotients are 
equal; that is, another equation is obtained. [Division Law~\ 

EXERCISES 

Find the value of the unknown numbers, doing all you 
can orally: 

1. 2 x + 3 = 9. 12. 5 r + ^ 2 - = 13|. 

2. 3x + 4 = 16. 13. 14 k + 7 = 79. 

3. 2 a + 5 = 17. 14. 3 e + 4j = 9. 

4. 3*4-7 = 28. 15. 15x4-0.5 = 26. 

5. 5 r 4- 7 = 62. 16. 11 m + J = ^ 

6. 9s 4- 21 = 93. 17. 1.3y + 3 = 16. 

7. 2y + l=S. 18. 11 y = 33. 

8. 5 y 4- 3 = 15. 19. 1.1 x = 121. 

9. 4x4-3.2=15.2. 20. 2.3x + 4 = 50. 

10. 6^4-4 = 49. 21. 6.3 z 4- 2.4 = 15. 

11. 9e + 8 = 116. 22. 5.3x4-0.34 = 2.99. 

5. Addition Law. In Fig. 3 the apparatus is so arranged 
that the 2-pound weight attached to the string which 
passes over the pulley pulls upward on the bar at B. This 



THE EQUATION 



-Pulley 



arrangement makes the problem different from the two 
which we have considered. If there were no pulley attach- 
ment, the weight pulling downward in the left pan would 
be 5 a; pounds. Since there is a lifting force of 2 Ib. at B, 
the downward pulling force in the left pan is 2 Ib. less 
than 5 x pounds, or 5 x 2 pounds ; this balances the 18 Ib. 
in the right pan. Hence the equation which describes the 
situation in Fig. 3 
is 5z-2 = 18. li- 
the string be cut so 
as to remove the up- 
ward pull of 2 Ib., 
then a 2-pound 
weight must be 
added to the right 
pan to keep the 
scales balanced, for 
removing the up- 
ward pull of 2 Ib. 
gave us a down- 
ward pull in the left pan of 5 a? pounds. This is 2 Ib. more 
than we had with the pulley attached, hence the necessity 
of adding 2 Ib. to the right pan. 

By the use of the equation the preceding discussion may 
take the following brief form: 

. . _ o _ i o / This expresses the 
o \. original conditions. 

M 

Adding 2 to both members, 




FIG. 3. IN THIS CASE THE SCALES ILLUSTRATE 
THE ADDITION LAW 



5 x 



= 20 



Dividing both members by 5, 



This problem illustrates the principle that if the same 
number is added to both members of an equation, the sums are 
equal ; that is, another equation is obtained. [Addition Law\ 



6 GENERAL MATHEMATICS 

EXERCISES 

Find the value of the unknown number in each problem, 
doing all you can orally : 

1. x -5 = 10. 12. 12m -35 = 41. 

2. 2 a; -15 = 13. 13. 9c- 3.2 = 14.8. 

3. 3 a; -12 = 13. 14. 7 * - 2 = 5.7. 

4. 3.r-8 = 17. 15. 14 A- - 5 = 21. 

5. 12 y - 4 = 46. . 16. 2 y - 3.1 = 3.2. 

6. 4 1 - 16 = 16. 17. 0.5 x - 3 = 4.5. 

7. 19,--4i = 14f 18. 2 x -1=61 

8. lly-9 = 79. 19. 3cc-9l = 17.r>. 

9. 56- - 0.1 = 0.9. 20. 9.7- -7.5 = 73.5. 

10. 4y-f = 7j. 21. 1.5 a; -3 = 1.5. 

11. 7*- 4 = 26. 22. 1.6 x- 1.7 = 1.5. 

6. Solving an equation ; check ; root. The process of 
finding the value of the unknown number in an equation 
is called solving the equation. To illustrate : 

Let y + 3 = 8 be the equation. 

"3 = 3 
Then y 5, and the equation is said to be solved. 

To test-, or check, the correctness of the result replace 
the unknown number in the original equation by 5, ob- 
taining 5 + 3 = 8. Since both members of the equation 
reduce to the same number, the result y = 5 is correct. 

When a number is put in place of a literal number it 
is said to be substituted .for the literal number. 

When both sides of an equation reduce to the same 
number for certain values of the unknown number, the 
equation is said to be satisfied. Thus, 3 satisfies the equa- 
tion y + 2 = 5. 



THE EQUATION 7 

A number that satisfies an equation is a root of the 
equation. 

Thus, 5 is a root of the equation z + 3 = 8. 

HISTORICAL NOTE. The word "root " first appears in the algebra 
of Mohammed ibn Musa Abu Jafar Al-Khwarizmi (about A.D. 830). 
The root of an equation (like the root of a plant) is hidden until 
found. See Ball's "A Short History of Mathematics," p. 163. 

EXERCISES 
Solve the following equations and check the results : 

1. 5y+3=18. 5. 26 + 2.7-1.3=11.4. 

2. 7z-4=17. 6. 7x-3x + 3.1 = 7.1. 

3. 2 a; -1.3 = 2.7. 7. 5* +14 -9 =15. 

4. 3a + 4.5=7.5. 8. 7m 3f = 3j. 

7. Terms; monomial; order of terms. The parts of an 
expression separated by plus (+) and minus ( ) signs are 
called the terms of a number. Thus, 2 a and 3 b are the 
terms of the number 2 a-\- 3 b. A one-term number is called 
a monomial. 

EXERCISES 
1.8-7+2 = ? 4. 8x7x + 2x = ? 

2.8 + 2-7=? 5. 8x + 2x-7x = ? 

3.2 + 8-7=? 6. 2z + 8z lx = ? 

These problems illustrate the principle (to be discussed 
more fully later) that the value of an expression is un- 
changed if the order of its terms is changed, provided 
each term carries with it the sign at its left. If no sign 
is expressed at the left of the first term of an expression, 
the plus sign is understood. 

8. Similar and dissimilar terms. Terms which have a 
common literal factor, as 2 r, 3 x, and 5 x, are similar terms. 
Their sum is a one-term expression ; namely, 10 a;. When 



8 GENERAL MATHEMATICS 

terms do not have a common literal factor, as 2 x and 
3 y, they are called dissimilar terms. Algebraic expressions 
are simplified by combining similar terms. Combining 
similar terms in either the right or the left member of an 
equation gives us the same equation in simpler form. 

HISTORICAL XOTE. The word "algebra" first appears about 
A.D. 830 in an Arabian work called " Al-jebr wa'1-mukabala," written 
by Al-Khwariznii. "Al-jebr," from which "algebra" is derived, may 
be translated by the restoration and refers to the fact that the same 
number may be added to or subtracted from both sides of the 
equation; "al-mukabala" means the process of comparison, and some 
writers say it was used in connection with the combination of 
similar terms into one term. 

The mathematical interest of the Arabs ran high. In the seventh 
century religious enthusiasm had banded these nomadic tribes 
into a conquering, nourishing nation. Enormous fortunes demanded 
mathematical manipulation. Cantor cites a rumor of a merchant 
whose annual income was about seven million dollars and a Christian 
doctor of medicine whose annual income was about fifty thousand. 
These fortunes gave them the necessary leisure time for culture 
and learning. Among the many books translated was the Greek 
geometry, Euclid's " Elements." See Ball's K A Short History of 
Mathematics," p. 162, and Miller's "Historical Introduction to 
Mathematical Literature," p. 83. 

EXERCISES 

Solve the following equations and check the results : 

1. 2./- -7 =x + 3. 

Subtract x from both members and proceed as usual. 

2. 3x + 2 = x + S. 

3. 5 i x 3x + 2-x 2 = 2x + 

4. 16y-8y + 3y-2 = 5y 

5. 20-4.r = 38 -10 a-. 

6. 5.T -j-3 05 = .r +18. 



THE EQUATION 9 



7. 7 > +18 + 3 r = 32 + 2 r - 2. 

8. 18 + 6 s + 30 + 6 s = 4 s + 8 + 12 + 3 * + 3 + s + 29. 

9. 25 y + 20 - 7 y - 5 = 56 - 5 // + 5. 

^J 

9. Multiplication Law. Solve for x: - = 5. 

A 

This problem offers a new principle. If we translate it 
into an English sentence, it reads as follows : One half of 
what unknown number equals 5 ? If one half of a num- 
ber equals 5, all of the number, or twice as much, equals 
2 times 5, or 10. The problem may be solved as follows: 



Multiplying both members by 2, 

2 x = 2 -x 5. 



x = 10. 

The principle involved here may be further illustrated 
by the scales, for if an object in one pan of a scales will 
balance a 5-pound weight in the other, it will be readily 
seen that three objects of the same kind would need 15 Ib. 
to balance them. This may be expressed in equation form 
as feitews : _ = 

X U. 

Mujtiplyijig both members by 3, 

3 x = 15. 

v. 

From the preceding' discussion it is evident that if 
both members of an equation are multiplied by the same 
number, the products are equal: that is, another equation 
is obtained. [Multiplication Law\ 



10 GENERAL MATHEMATICS 

This multiplication principle is convenient when an 
equation contains fractions. It enables us to obtain a 
second equation containing no fraction but containing the 
same unknown number. To illustrate this: 

Let \ .- = 7. 

Multiplying both members by 3, 

3 x i- x = 3 x 7. 

Reducing to simplest form, x = 21. 

ORAL EXERCISES 

Find the value of the unknown number in each of the 
following equations : 



tn /> & IT 



The preceding list of problems shows that it is desirable 
to multiply both members of the equation by some number 
that will give us a new equation without fractions. The same 
principle holds when the equation contains two or more 
fractions whose denominators are different, as is illustrated 
by the following problem : 

Find x if 7-| = 2. 
4 5 

Solution. - - - = 2. 

4 5 

90 r 90 r 

Multiplying by 20, = 40. 

Simplifying, 5 x 4 x 40 ; 

whence x = 40. 



THE EQUATION 11 

The fact that 4 and 5 will divide integrally (a whole 
number of times) into the numerators gives us a new equa- 
tion without fractions. Obviously there are an unlimited 
number of numbers (for example, 40, 60, 80, etc.) which 
we could have used, but it was advantageous to use the 
smallest number in which 4 and 5 are contained integrally ; 
namely, the least common multiple of 4 and 5, which is 
20. The object of this multiplication is to obtain an equa- 
tion in which the value of the unknown number is more 
easily found than in the preceding one. This discussion 
may be summarized by the following rule: 

If the given equation contains fractions, multiply every term 
in both members by the least common multiple {L.C.M.} of 
the denominators in order to obtain a new equation which does 
not contain fractions. 

EXERCISES 

Find the value of the unknown number in each problem, 
and check : 



2 ^ + ^-9 a 2x , x x , l 

<S i . <? H. - -f- = - + 

54 96 18 3 



4. it + J y = 6. ' K 

11 ^ 

5- y - I y = 7. 6 



12 GENERAL MATHEMATICS 

10. Definition of the equation; properties of the equation. 

The foregoing problems were used to show that an equa- 
tion iif a statement that two numbers are equal. It indicates 
that two expressions stand for the same number. It may 
be regarded as an expression of balance of values be- 
tween the numbers on the two sides of the equality sign. 
Some unknown number which enters into the discussion 
of the problem is represented by a letter. An equation 
is written which enables one to find the value of that 
unknown number. 

An equation is like a balance in that the balance of 
value is not disturbed so long as like changes are made 
on both sides. Thus, in equations we may add the same 
number to both aides, or subtract the same number from 
both sides, or we may multiply or divide both sides by the 
same number (except division by zero); the equality is 
maintained during all these changes. 

On the other hand, the equality is destroyed if more is 
added to or subtracted from one side than the other or if 
one side is multiplied or divided by a larger number than 
is the other side. 

11. Translation of an equation. The list of problems in 
the preceding exercises may appear abstract in the sense 
that the equations do not appear to be connected in any 
way either with a concrete situation of a verbal problem, 
or with our past experience. However, such a list need 
not be regarded as meaningless. Just as an English sen- 
tence which expresses a number relation may be written 
in the " shorthand " form of an equation, so, conversely, 
an equation may be translated into a problem ; for ex- 
ample, the equation 3x + 5 = 2x + 2Q may be interpreted 
as follows : Find a number such that 3 times the num- 
ber plus 5 equals 2 times the number plus 20. The 



THE EQUATION 13 

equation .c 3 = 5 may be regarded as raising the question 
What number diminished by 3 ' equals 5 ? Or, again, 
21 x + x + 2x + x = 140 may be considered as the trans- 
lation of the following problem: What is the altitude of 
a rectangle whose base is 21 times as long as the altitude 
and whose perimeter is 140 ft. ? 

EXERCISES 

State each of the following in the form of a question or a 
verbal problem : 

1. 05-6 = 3. 7. 7r-2 = 6r + 8. 

2. 2a- -1 = 10. 8. 5.2 x- 3 = 4.1 x+ 1.4. 

3. 9 k -10 = 87. 9. 3* =12. 

4. ly + 8 = 112. 10. 4* =16. 

5. 7s- 3 = 81. 11. 2x + 3x + 4x = 18. 

6. 3x + 2 = 2x + :;. 12. c = 2 T 2 rl. 

12. Drill in the "shorthand" of algebra. The following 
exercises give practice in translating number expressions 
and relations from verbal into symbolical language: 

1. Consecutive numbers are integral (whole) numbers which 
follow each other in counting ; thus, 17 and 18, 45 and 46, are 
examples of consecutive numbers. Begin at s -f 5 and count 
forward. Begin at x + 3 and count backward. Give four con- 
secutive integers beginning with 18 ; ending with 18 ; beginning 
with x; ending with x. Give two consecutive even integers 
beginning with 2 x. Give two consecutive even integers end- 
ing with 2c. 

2. The present age, in years, of a person is denoted by x. 
Indicate in symbols the following : (a) the person's age fourteen 
years ago ; (b) his age fourteen years hence ; (c) his age when 
twice as old as now ; (d) 60 decreased by his age ; (e) his age 
decreased by 60 ; (f) his age increased by one half his age. 



14 GENERAL MATHEMATICS 

3. A boy has a marbles and buys b more. How many has 
he ? What is the sum of 'a and b ? 

4. A boy having b marbles loses c marbles. How many 
has he? 

5. (a) The home team made 8 points in a basket-ball game 
and the visiting team made 3 points. By how many points did 
the home team win ? (b) If the home team scored h points 
to the visitors' n points, by how many points did the home 
team win ? (c) Substitute numbers for h in the last question 
that will show the defeat of the home team, (d) If h = 5, what 
must be the value of n when the game is a tie ? 

6. What is the 5th part of n ? f of y ? f of t ? 

7. Two numbers differ by 7. The smaller is s. Express 
the larger number. 

8. Divide 100 into two parts so that one part is . 

9. Divide a into two parts so that one part is 5. 

10. The difference between two numbers is d and the larger 
one is I. Express the smaller one. 

11. What number divided by 3 will give the quotient a? 

12. A man's house, worth h dollars, was destroyed by fire. 
He received i dollars insurance. What was his total loss ? 

13. A is x years old and B lacks 5 yr. of being three times 
as old. Express B's age. 

14. A has m ties and B has n ties. If A sells B 5 ties, how 
many will each then have ? 

15. A man has d dollars and spends c cents. How many 
cents does he have left ? 

16. A room is I feet long and w feet wide. How many feet 
of border does such a room require ? 

17. The length of a rectangle exceeds its width by c feet. 
It is w feet wide, (a) State the length of each side, (b) Find 
the distance around the rectangle. 



THE EQUATION 15 

18. What is the cost of 7 pencils at c cents each ? 

19. What is the cost of 1 sheet of paper if b sheets can be 
bought for 100? 

20. It takes 'two boys 5 da. to make an automobile trip. 
What part can they do in 1 da. ? If it takes them d days, 
what part of the trip do they travel in 1 da. ? 

21. If a man drives a car at the rate of 31 mi. per hour, 
how far can he drive in 3 hr. ? in 5 hr. ? in li hours ? 

22. If a man drives n miles in 3 hr., how many miles does 
he go per hour ? 

23. A tank is filled by a pipe in m minutes. How much of 
the tank is filled in 1 min. ? 

24. The numerator of a fraction exceeds the denominator 
by 3. (a) Write the numerator, (b) Write the fraction, 
(c) Head the fraction. 

25. A pair of gloves costs d dollars. What is the cost if the 
price is raised 70? if lowered 70? 

26. Write the sum of x and 17 ; of 17 and x. Write the 
difference of x and 17; of 17 and x. 

27. A class president was elected by a majority of 7 votes. 
If the unsuccessful candidate received k votes, how many votes 
were cast ? 

13. Algebraic solution. Many problems may be solved 
by either arithmetic or the use of the equation. When 
the solution of a problem is obtained by the use of the 
equation, 'it is commonly called an algebraic solution. 

The following problems illustrate the important steps 
in the algebraic solution of a problem. By way of contrast 
an arithmetic solution is given for the first problem. 

1. Divide a pole 20 ft. long into two parts so that one part 
shall be four times as long as the other. 



16 GEXKKAL .MATHEMATICS 

ARITHMETICAL SOLUTION 
The shorter part is a certain length. 
The longer part is four times this length. 
The whole pole is then five times as long as the shorter part. 
The pole is 20 ft. long. 
The shorter part is \ of 20 ft., or 4 ft. 
The longer part is 4 x 4 ft., or 16 ft. 
Hence the parts are 4 ft. and 16 ft. long respectively. 

ALGEBRAIC SOLUTION 

Let n = number of feet in the shorter part. 

Then 4 n = number of feet in the longer part, 

and n + 4 n, or 5 n length of the pole. 

Then ."> n = '20. 

n = 4. 
4 n = 10. 

Hence the parts are 4 ft. and 1(J it. long respectively. 
/ 
2. A rectangular garden is three times as long as it is wide. 

It takes 80 yd. of fence to inclose it. Find the width and 

length. 

ALGEBRAIC SOLUTION 

Let x = number of feet in the width. 

Then .3 x = number of feet in the length, 

and ./ + :> x + x + 3 x = distance around the garden. 

Then 8 x = 80. 

x = 10. 
3 x = 30. 
Hence the width is 10 yd. and the length is 30 yd. 

14. The important steps in the algebraic solution of verbal 
(or story) problems. Before proceeding to the solution of 
more difficult problems it is important that we organize the 
steps that are involved. The preceding list of problems illus- 
trates the following method for solving a verbal problem : 

(a) In every problem certain facts are given as known 
and one or more as unknown and to be determined. Read 
the problem so as to get these facts clearly in mind. 



THE EQUATION 17 

(b) In solving the problem denote one of the unknown 
numbers by some symbol, as y. 

(c) Then express all the given facts in algebraic lan- 
guage, using the number y as if it, too, were known. 

(d) Find two different expressions which denote the 
same number and equate them. (Join by the sign of 
equality (=).) 

(e) Solve the equation for the value of the unknown 
number. 

(f) Check the result by re-reading the problem, substi- 
tuting the result in the conditions of. the problem to see 
if these conditions are satisfied. Note that it is not suffi- 
cient to check the equation, for you may have written the 
wrong equation to represent the conditions of the problem. 

EXERCISES 

With the preceding outline of method in mind, solve the 
following problems, and check : 

PROBLEMS INVOLVING NUMBER RELATIONS 

1. If six times a number is decreased by 4, the result is 26. 
Find the number. 

2. If four fifths of a number is decreased by 6, the result 
is 10. Find the number. 

3. The sum of three numbers is 12Q. The second is five 
times the first, and the third is nine times the first. Find 
the numbers. 

4. The sum of three numbers is 360. The second is four- 
teen times the first, and the third is the sum of the other two. 
Find the numbers. 

5. Seven times a number increased by one third of itself 
equals 44. Find the number. 



18 GENERAL MATHEMATICS 

6. The following puzzle was proposed to a boy : " Think 
of a number, multiply it by 4, add 12, subtract 6, and divide 
by 2." The boy gave his final result as 13. What was his 
original number ? 

7. The sum of one half, one third, and one fourth of a 
number is 52. What is the number ? 

CONSECUTIVE-NUMBER PROBLEMS 

8. Find two consecutive numbers whose sum is 223. 

9. Find three consecutive numbers whose sum is 180. 

10. Find two consecutive odd numbers whose sum is 204. 

* 

11. Find three consecutive even numbers whose sum is 156. 

12. It is required to divide a board 70 in. long into five parts 
such that the four longer parts shall be 1", 2", 3", and 4" 
longer respectively than the shortest part. Find the lengths 
of the different parts. 

13. A boy in a manual-training school is making a bookcase. 
The distance from the top board to the bottom is 4 ft. 7 in., 
inside measure. He wishes to put in three shelves, each 1 in. 
thick, so that the four book spaces will diminish successively 
by 2 in. from the bottom to the top. Find the spaces. . 

PROBLEMS INVOLVING GEOMETRIC RELATIONS 

14. The length of a field is three times its width, and the 
distance around the field is 200 rd. If the field is rectangular, 
what are the dimensions ? 

15. A room is 15 ft. long, 14 ft. wide, and the walls contain 
464 sq. ft. Find the height of the room. 

16. The perimeter of (distance around) a square equals 64 ft. 
Find a side. 

NOTE. Such geometric terms as "triangle," "rectangle," "square," 
etc., as occur in this list of problems (14-24), are familiar from arith- 
metic. However, they will later be defined more closely to meet 
other needs. 



THE EQUATION 19 

17. Find the sides of a triangle if the second side is 3 ft. 
longer than the first, the third side 5 ft. longer than the first, 
and the perimeter is 29 ft. 

18. Find the side of an equilateral (all sides equal) triangle 
if the perimeter is 21f ft. 

19. The perimeter of an equilateral pentagon (5-sided figure) 
is 145 in. Find a side. 

20. The perimeter of an equilateral hexagon (6-sided figure) 
is 192 ft. Find a side. 

21. Find the side of an equilateral decagon (10-sided figure) 
if its perimeter is 173 in. 

22. What is the side of an equilateral dodecagon (12-sided 
figure) if its perimeter is 288 in. ? 

23. A line 60 in. long is divided into two parts. Twice the 
larger part exceeds five times the smaller part by 15 in. How 
many inches are in each part ? 

24. The perimeter of a quadrilateral A BCD (4-sided figure) 
is 34 in. The side CD is twice as long, as the side AB; the 
side AD is three times as long as CD; the side BC equals the 
sum of the sides AD and CD. Find the length of each side. 

MISCELLANEOUS PROBLEMS 

25. Divide $48,000 among A, B, and C so that A's share 
may be three times that of B, and C may have one half of 
what A and B have together. 

26. The perimeter of a rectangle is 132 in. The base is 
double the altitude. Find the dimensions of the rectangle. 

27. A and B own a house worth $16,100, and A has invested 
twice as much capital as B. How much has each invested ? 

28. A regulation football field is 56|^ yd. longer than it is 
wide, and .the sum of its length and width is 163^ yd. Find 
its dimensions. 



20 GENERAL MATHEMATICS 

29. A man has four times as many chickens as his neighbor. 
After selling 14, he has 3^ times as many. How many had 
each before the sale ? 

30. In electing a president of the athletic board a certain 
high school cast 1019 votes for three candidates. The first 
received 143 more than the third, and the second 49 more 
than the first. How many votes did each get ? 

31. A boy has $5.20 and his brother has $32.50. The first 
saves 200 each day and the second spends 100 each day. In 
how many days will they have the same amount ? 

32. One man has seven times as many acres as another. 
After the first sold 9 A. to the second, he had 36 A. more 
than the second then had. How many did each have before 
the sale ? 

33. To find the weight of a golf ball a man puts 20 golf 
balls into the left scale pan of a balance and a 2-pound weight 
into the right; he finds that too much, f but the balance is 
restored if he puts 2. oz. into the left scale pan. What was 
the weight of a golf ball ? 

34. The number of representatives and senators together 
in the United States Congress is 531. The number of repre- 
sentatives is 51 more than four times the number of senators. 
Find the number of each. 

35. A boy, an apprentice, and a master workman have the 
understanding that the apprentice shall receive twice as much 
as the boy, and the master workman four times as much as 
the boy. How much does each receive if the total amount 
received for a piece of work is $105 ? 

36. A father leaves $13.000 to be divided among his three 
children, so that the eldest child receives $2000 more than 
the second, and twice as much as the third. What is the 
share of pa oh ? 



THE EQUATION 21 

37. A fence 5 ft. high is made out of 6-inch boards running 
lengthwise. The number of boards necessary to build the 
fence up to the required height is 5 and they are so placed 
as to leave open spaces between them. If each of these open 
spaces, counting from the bottom upwards, is half of the one 
next above it, what must be the distances between the boards ; 
that is, what will be the width of each of the open spaces ? 

38. I paid $8 for an advertisement of 8 lines, as follows : 
240 a line for the first insertion, 100 a line for each of the 
next five insertions, and 20 a line after that. Firid the 
number of insertions. 

39. The annual income of a family is divided as follows : 
One tenth is used for clothing, one third for groceries, and one 
fifth for rent. This leaves $660 for other expenses and for 
the savings account. How much is the income ? 

15. Axioms. Thus far we have used the four following 
laws in solving equations: 

I. If the same number be added to equal numbers, the 
sums are equal. [Addition Law] 

II. If the same number be subtracted from equal numbers, 
the remainders are equal. [Subtraction Law] 

III. If equal numbers be multiplied by the same number, 
the products are equal. [Multiplication Law] 

IV. If equal numbers be divided by the same number 
(excluding division by zero), the quotients are equal. [Division 
Law] 

Statements like the four laws above, when assumed to 
be true, are called axioms. Usually axioms are statements 
so simple that they seem evident. A simple illustration 
is sufficient to make clear the validity of the axiom. For 
example, if two boys have the same number of marbles 
and 3 more are given to each, then our experience tells 



22 GENERAL MATHEMATICS 

us that again one boy would have just as many as the 
other. This illustrates the validity of the addition axiom. 
Hereafter the preceding laws will be called Axioms I, II, 
III, and IV respectively. 

16. Axiom V. In this chapter we have also made fre- 
quent use of another axiom. In solving a verbal problem 
we obtained the necessary equation by finding two expres- 
sions which denoted the same number and then we equated 
these two expressions. This step implies the following 
axiom : 

If two numbers are equal to the same number (or to equal 
numbers'), the numbers are equal. [Equality Axiom] 

Illustrate the truth of Axiom V by some familiar 
experience. 

The following exercises test and review the axioms. 

EXERCISES 

Solve the following equations, and check. Be able to state 
at every step in the solution the axiom used. 

1. 12 #-15 = 30. 

Solution. 12 1 - 15 =30 

15 = 15 (Axiom I) 

Adding 15 to both members, 12 t = 45 

Dividing both sides by 12, / = f f, or 3. (Axiom IV) 

y y 1 

2. - + - = - 

24 3 

Solution. Multiplying both sides of the equation by the least 
common multiple of the denominators, that is, by 12, 

12 y , 12 y 12 

-2^ + -^ : = y (Axiom III) 

Then 6 y + 3 y = 4. 



THE EQUATION 23 

By reducing the fractions of the first equation to lowest terms 
we obtain the second equation, which does not contain fractions. 

Combining similar terms, 9 y 4. 
Dividing both sides of the equation by 9, 

y = f. (Axiom IV) 

3. 12a-+13 = 73. 8. 17s- 3s + 16s =105. 

4. 18r-12r = 33. 9. 17x + 3x - 9x = 88. 

5. 21^+15=120. 10. 16m + 2m 13m = 22%. 

6. 28* -9 = 251. 11. 202y-152// + 6?/ = 280. 

7. 20y+ 2y-18y = 22. 12. 3.4 x 1.2 x + 4.8 x = 70. 

13. 3.5 y + 7.6 ?/- 8.6y=15. 

14. 5.8 m 3.9 m + 12.6 m = 58. 

15. 6 > - 3.5 > + 5.5 r = 68. 

16. 3.41 x + 0.59 x - 1.77 x = 22.3. 

17. 8 y 4.5 y + 5.2 y = 87. 

18. 2s +7s - 3s- 6 = 24. 

19. lx = 6. 27. ^ + |_| = 36. 

20. J* = 2. 28. 15 = 3sc-3. 

21. fa; = 6. Solution. Adding 3 to both 

members, 

22. fa; = 25. 18 = 3*. 

^ jj. J)ividing both members by 3, 

23- 3 + 2 =1 - 6=x. 

Note that in the preceding 

24. + = 3. problem the unknown appears 

in the right member. 

25. ^ + ^ = 8. 29. 17=2^-3. 

30. 

26.-^ = 6. 31. 



24 GENERAL MATHEMATICS 

32.^ = 4. 33.^ = 5. 

x b 

Solution. Multiplying both _ . 16 _ 

members by #, 3 x 

1Q = ix. Multiply both members by 3 x. 

Hence 4 = x. _ _ 3 

<5o. - = 1. 

4 
Note that in the preceding 

problem the unknown occurs in ,g 13 _ . 

the denominator. 2 

SUMMARY 

17. This chapter has taught the meaning of the follow- 
ing words and phrases : equation, members of an equation, 
equation is satisfied, substituting, check, root of an equa- 
tion, verbal problem, algebraic solution of a verbal problem, 
term, monomial, literal number, similar terms, dissimilar 
terms, order of terms, and axiom. 

18. Axioms. In solving equations the following axioms 
are used : 

I. If the same number or equal numbers be added to equal 
numbers, the sums are equaL [Addition Axiom] 

II. If the same number or equal numbers be subtracted from 
equal numbers, the remainders are equal. [Subtraction Axiom] 

III. If equal tno/ihcrs be divided by equal numbers (exclud- 
ing division by zero), the quotients are equal. [Division Axiom] 

IV. If equal numbers be multiplied by the same number or 
equal numbers, the products are equal. [Multiplication Axiom] 

V. If two numbers are equal to the same number or to equal 
numbers, the numbers are equal. [Equality Axiom] 

19. If an equation contains fractions, a second equation 
involving the same unknown may be obtained by multi- 
plying every term of the given equation by the L.C.M. 



THE EQUATION 25 

of the denominators and then reducing all fractions to 
integers. The solution of this equation is more readily 
obtained than, that of the given equation. 

20. The equation is a convenient tool for solving 
problems. In solving a verbal problem algebraically ob- 
serve the following steps: 

(a) Use a letter for the unknown number called for in 
the problem. Often the last sentence suggests the most 
convenient choice. 

(b) Express the given facts in terms of the unknown 
(provided, of course, that these facts are not definite arith- 
metical numbers). 

(c) Obtain two expressions for the same number and 
equate them. (This gives us the equation.) 

(d) Solve the equation. Check the result by substitut- 
ing in the conditions of the problem. 



CHAPTER II 

LINEAR MEASUREMENT. THE EQUATION APPLIED TO 

LENGTH 1 

21. Length, the important characteristic of lines. If in 
drawing an object we lay the ruler on a sheet of paper and 
pass a sharpened pencil as near the edge as possible, thus 
obtaining what is familiar to us as a straight line, we are at 
once concerned with the length of a part of the straight line 
drawn. In fact, length is the important characteristic of a 
line. In an exact sense a line has length only, not width nor 

thickness. Thus, A ^ 

the edge of a table 

FIG. 4. A LINE SEGMENT 
has length only ; 

the thickness and width of a crayon line are neglected ; the 
wide chalk marks on a tennis court are not boundary 
lines, but are made wide to help us see the real boundary 
lines, which are the outside edges of the chalk marks. 

The part of a line whose length we wish to determine 
is a line segment or, briefly, a segment, as AB in Fig. 4. 

A line segment has a definite beginning point and a 
definite ending point. The word "point" is used to mean 
merely position, not length, breadth, nor thickness. The 
position of a point is shown by a short cross line ; that is, 
a point is determined by two intersecting lines. 

1 The pupil should now provide himself with a ruler one edge of which 
is graduated to inches and fractional parts of an inch and the other to 
units of the metric scale. He should also obtain a pair of compasses 
and some squared paper ruled to the metric scale. 

26 



LINEAR MEASUREMENT 27 

EXERCISE 

Give examples of line segments that can be seen in the 
classroom. 

22. Measurement of length. In Fig. 5 the length of the 
line segment AB is to be determined. One edge of your 
ruler is graduated (divided) into inches and fractions of 
an inch as is shown in Fig. 5. Place the division marked 
zero on your ruler at A, with the edge of the ruler along 

A B 

i f 



FIG. 5. How A LINE SEGMENT MAT BE MEASURED 

the segment AB, and read the number of inches in the line 
segment AB; that is, find what reading on the ruler is 
opposite the point B. 

In the preceding problem we compared the unknown 
length of the line segment AB with the well-established 
and well-known segment, the inch, and found the line 
segment to be 21 times as long as the inch. Hence the 
length of the line segment is 2^ in. When we determine 
the length of a line segment we are measuring the line 
segment. The segment with which we compare the given 
line is called a unit segment or a unit of measurement. Hence 
to measure a line segment is to apply a standard unit seg- 
ment to it to find out how many times the unit segment is 
contained in it. 

EXERCISES 

1. Draw a line segment and express its length in inches. 

2. Measure the length of your desk in inches. 

3. Measure the width of your desk in inches. 



28 (rEXEKAL MATHEMATICS 

23. Different units of length. The most familiar 
inents for measuring are the foot rule, the yardstick, and 
several kinds of tape lines. The fractional parts of the unit 
are read by means of a graduated scale engraved or stamped 
on the standard unit used. In Fig. 6, below, is shown a 
part of a ruler. The upper edge is divided into inches 
and fractional parts of an inch. What is the length of the 
smallest line segment of the upper edge ? 

The lower edge of the ruler is divided into units of 
the metric (or French) scale. This system is based on the 



I ' i ' I ' I ' ' t ' i ' i ' ' i ' i ' i ' i ' M i ' i ' M 1 1 1 1 1 ' M M i f m 1 1 1 1 ' 1 1 1 

O INCH i 2 3 

O 1 2 3 4 

I .. CENTIMETERS | . , . ..| .... I. 

u 

' I-> 

iCm. 
FIG. 6. PART OF A RULER, SHOWING DIFFERENT UNITS OF LENGTH 

decimal system and is now very generally used in scien- 
tific work in all countries. The standard unit is called the 
meter (m.). It is divided into 1000 equal parts called milli- 
meters (mm.). In the figure above, the smallest division 
is a millimeter. Ten millimeters make a centimeter (cm.). 
In the figure, AB is one centimeter in length, and you will 
note that this is about two fifths of an inch. Ten centi- 
meters make a decimeter (dm.) (about 4 in.), and ten deci- 
meters make a meter (39.37 in., or about 1.1 yd.). We may 
summarize these facts in the following reference table : 

f 1 millimeter = 0.03937 inches 
\linch = 2.54 centimeters 

10 millimeters = 1 centimeter (0.3937 in., or nearly J in.) 
10 centimeters = 1 decimeter (3.937 in., or nearly 4 in.) 

10 decimeters = 1 meter (39.37 in., or nearly 3-^ ft.) 



LINEAR MEASUREMENT 29 

24. Advantages of the metric system. One of the advan- 
tages of the system is that the value of the fractional part 
of the meter is more apparent than a corresponding deci- 
mal part of a yard. Thus, if we say a street is 12.386 yd. 
wide, the decimal 0.386 tells us nothing about the smaller 
divisions of a yardstick that enter into this number. At 
best we would probably say that it is something over one 
third of a yard. On the other hand, if we say that a road 
is 12.386 m. wide, we know at once that the road is 12 m. 
3 dm. 8 cm. 6 mm. wide. This last statement is far more 
definite to one who has had .a little practice with the 
metric system. 

Obviously the advantages of the metric system lie in 
the fact that ten line segments of any unit are equal to 
one of the next larger. In contrast to this fact the multi- 
pliers .of our system, though they may seem familiar, are 
awkward. Thus, there are 1 2 in. in a foot, 3 ft. in a yard, 
5^ yd. in a rod, 1760yd. in a mile, etc. 

HISTORICAL NOTE. It is probable that most of the standard units 
of length were derived from the lengths of parts of the human body 
or other equally familiar objects used in measuring. Thus, we still 
say that a horse is so many hands high. The yard is supposed to 
have represented the length of the arm of King Henry I. Nearly all 
nations have used a linear unit the name of which was derived from 
their word for foot. 

During the French Revolution the National Assembly appointed 
a commission to devise a system that would eliminate the inconven- 
ience of existing weights and measures. The present metric system 
is the work of this commission. 

This commission attempted to make the standard unit one ten- 
millionth part of the distance from the equator to the north pole 
measured on the meridian of Paris. Since later measurements have 
raised some doubt as to the exactness of the commission's determina- 
tion of this distance, we now define the meter not as a fraction of 
the earth's quadrant, but as the distance, at the freezing temperature, 



30 GENERAL MATHEMATICS 

between two transverse parallel Jim-.* ruled on a bar of platinum-iridium 
which is kept at the International Butqau of Weight.* and Measures, at 
Sevres, near Paris. 

25. Application of the metric scale. This article is in- 
tended to give practice in the use of the metric system. 

EXERCISES 

1. With a ruler whose edge is graduated into centimeters 
measure the segments AB and CD in Fig. 7. 






FIG. 7 

2. Measure the length and width of your desk with a centi- 
meter ruler. Check the results with those of Exs. 2 and 3, 
Art. 22. 

3. Estimate the length of the room in meters and then 
measure the room with a meter stick. If a meter stick is not 
available, use a yardstick and translate into meters. 

4. Turn to some standard text in physics (for example, 
Millikan and Gale, pp. 2 and 3) and report to the class on the 
metric system. 

5. Refer to an encyclopedia and find out what you can 
about the " standard yard " kept at Washington. 

26. Practical difficulty of precise measurement. In spite 
of the fact that measuring line segments is a familiar 
process and seems very simple, it is very difficult to meas- 
ure a line with a high degree of accuracy. The following 
sources of error may enter into the result if we use a 
yardstick : (1) the yardstick may not be exactly straight ; 
(2) it may be a little too long or too short ; (3) it may 
slip a little so that the second position does not begin at 



LINEAR MEASUREMENT 



31 



the exact place where the first ended ; (4) the edge of 
the. yardstick may not always be along the line segment; 
(5) the graduated scale used for reading feet, inches, and 
fractional parts of inches may not be correct. Nor do we 
eliminate these errors by using other measuring devices. 
For example, a tape line tends to stretch, but contracts if 
wet, while a steel tape is affected by heat and cold. 

From the preceding discussion it is apparent that a 
measurement is always an approximation. The error can 
be decreased but never wholly eliminated. 

EXERCISES 

1. Suppose you have measured a distance (say the edge of your 
desk) with great care and have found it to be equal to 2 ft. 7f in. 
Is this the exact length of the desk ? Justify your answer. 

2. If you were to repeat the measurement with still greater 
care, making use of a finer-graduated scale, is it likely that 
you would find exactly the same result 

as before ? 

3. If you were asked to measure 
the length of your classroom, what 
would you use ? Why ? 




27. The compasses. A pair of com- 
passes (Fig. 8) is an instrument 
that may be used in measuring line 
segments. Since the use of com- 
passes greatly decreases some of the 
common errors in measuring that 
have been pointed out, and is con- 
sequently very useful in many forms of drawing which 
require a high degree of accuracy, it will be helpful if the 
student learns to use the compasses freely. 



FIG. 8. A PAIR OF 
COMPASSES 



32 GENERAL MATHEMATICS 

28. Measuring a line segment with the compasses. To 
measure the line segment AB in Fig. 9 with the compasses, 
place the sharp points of the compasses on A and B. Turn 



FIG. 9 

the screw which clamps the legs of the compasses. Then 
place the points on the marks of the ruler and count the 
number of inches or centimeters between them. 

EXERCISES 

1. With the compasses measure AB in Fig. 9, in inches. 

2. With the compasses measure AB in Fig 9, in centimeters. 

3. Multiply the result of Ex. 1 by 2.54. What do you observe ? 

4. Estimate the number of centimeters in the length of this 
page. Measure the length of this page with the compasses and 
compare with your estimate. 

29. Squared paper. Squared paper is another important 
device which is often useful in measuring line segments. 

Squared paper is ruled either A B 

to inches and fractions of an 
inch (used by the engineer) or 
to the units of the metric scale. 
A sample part of a sheet is 
shown in Fig. 10. The method 
of measuring with squared pa- 
per is practically the same as 
measuring with the compasses 
and ruler. Thus, to measure 
the line segment AB in Fig. 10 
place the sharp points of the compasses on A and B. 
Clamp the compasses. Place the sharp points on one of 



FIG. 10. HOAV A LINE SEGMENT 

MAY BE MEASURED BY THE USE 

OF SQUARED PAPER 



LINEAR MEASUREMENT 



33 



the heavy lines, as at E and D. Each side of a large 
square being 1 cm., count the number of centimeters in 
ED and estimate the remainder to tenths of a centimeter. 
Thus, in Fig. 10 the segment ED equals 2.9 cm. 

EXERCISES 

1. Draw a line segment and measure its length in centi- 
meters by the use of squared paper. 

2. Why do you find it convenient to place one of the sharp 
points of the compasses tyhere two heavy lines intersect ? 

3. What are the advantages of measuring with squared 
paper over measuring with a ruler ? 

30. Measuring a length approximately to two decimal 
places. By increasing the size of the unit of measurement 
we may express the result with approximate accuracy to 
two decimal places. In Fig. 11 let MN (equal to ten small 
units, or 2 cm., of A % 

the squared paper) ' 
be the unit. As 
before, lay off AB 
upon the squared 
paper with the 
compasses (CD in 
the new position). 
Now EF (a small 
unit) equals 0.1 of 
a unit, and 0.1 of EF equals 0.01 of the unit MN. Reading 
the units in the line CD, we are sure the result is greater 
than 2.7, for the crossing line is beyond that. But it is not 
2.8. Why? Now imagine a small unit, as EF, divided 
into tenths and estimate the number of tenths from the 
end of the twenty-seventh small unit to D. This appears 



N 



Unit 



FIG. 11 



GENERAL MATHEMATICS 



to be 0.4, but this is 0.04 of a unit;, hence CD equals 
2.74 units. This means that CD is 2.74 times as long as 
the line MN. Of course the 4 is only an approximation, 
but it is reasonably close. 



\ 



Kt- 



EXERCISES 

1. In Fig. 12 measure to two decimal places the segments 
CD, DE, and EC. Compare the results of your work with 

that of the other members of 

the class. 

2. Is the result obtained by 
the method of Art. 30 more 
accurate than the result ob- 
tained by using 1 cm. as a unit 
and claiming accuracy to only 

one decimal place ? 

FIG. 12 

31. Equal line segments. 

When the end points of one segment, as a in Fig. 13, coin- 
cide with (exactly fit upon) the ends of another segment, 
as 6, the segments a and b are said to be 
equal. This fact may be expressed by 
the equation a = b. 

32. Unequal segments; inequality. If 
the end points of two segments, as a and J, 

cannot possibly be made to coincide, the segments are said to 
be unequal. This is written a = b (read " a is not equal to 6"). 

The statement a 3= bis called . a | 

an inequality. In Fig. 14 seg- 
ment a is less than segment b 
(written a < i), and seg- 
ment c is greater than seg- 
ment b (written c > 6). 



I 1 

I b - 1 

FIG. 13. EQUAL LINE 
SEGMENTS 



V 



FIG. 14. UNEQUAL LINE SEGMENTS 



LINEAR MEASUREMENT 35 

In the preceding equation and inequalities we need to 
remember that the letters a, b; and c stand for the length 
of the segments. They represent numbers which can be 
determined by measuring the segments. 

33. Ratio. This article will show that ratio is a funda- 
mental notion in measurement. 

INTRODUCTORY EXERCISES 
(Exs. 1-4 refer to Fig. 15) 

1. Measure the segment a accurately to two decimal places. 

2. Measure the segment b 

accurately to two decimal | 1 

places. 

3. What part of b is a ? \ I 

4. Find the quotient of b FlG 15 
divided by a. 

The quotient of two numbers of the same kind is called 
their ratio. The ratio is commonly expressed as a fraction. 
Before forming the fraction the two quantities must be 
expressed in terms of the same unit; for example, the 
ratio of 2 ft. to 5 in. is the ratio of 24 in. to 5 in., that 
is, - 2 ^ 4 -. The unit of measure is 1 in. Obviously there is 
no ratio between quantities of different kinds ; for exam- 
ple, there is no ratio between 7 gal. and 5 cm. 

It should now be clear that every measurement is the 
determination of a ratio either exact or approximate. Thus, 
when we measure the length of the classroom and say it> is 
10 m. long, We mean that it is ten times as long as the 
standard unit, the meter ; that is, the ratio is. -L<>.. 

- - - - ' -. .- - - : 



36 



GENERAL MATHEMATICS 



EXERCISES 

1. The death rate in Chicago in a recent year was 16 to 1000 
population. Express this ratio as a fraction. 

2. An alloy consists of copper and tin in the ratio of 2 to 3. 
What part of the alloy is copper ? What part is tin ? 

3. A solution consists of alcohol and water in the ratio of 
3 to 6. What part of the solution is water ? 

4. The ratio of weights of equal volumes of water and cop- 
per is given by the fraction How many times heavier is 
copper than water ? 

5. Water consists of hydrogen and oxygen in the ratio of 
1 to 7.84. Express this ratio as a decimal fraction. 

34. Sum of two segments ; geometric addition. It is 

possible to add two line segments by the use of compasses. 
Thus, in Fig. 16 if the segment a is laid off on the number 
scale of squared paper from point A to point B and if in 
turn b is laid off on the same line from B to C, then the 



FIG. 16. GEOMETRIC ADDITION OF LINE SEGMENTS 

sum of these lines can be read off at once. In Fig. 16 the 
sum is 5.4 cm. The segment AC is the sum of a and b. 
Very often in construction work we are not concerned 
about either the length of the segments or their sum. In 
that case lay off the segments as above on a working line 
and indicate the sura of a and b as a +- b. Addition per- 
formed by means of the compasses is a geometric addition. 



LINEAR MEASUREMENT 



37 



FIG. 17 



EXERCISES 

1. In Fig. 17 find the sum of a, b, and c on a working 

line. Indicate the sum. 

. . 

2. In Fig. 17 add the seg- 

ments a, b, and c on the scale i - 1 

line of a sheet of squared c 

paper. Express the value of 
a + b + c in centimeters. 

3. On a working line draw 
one line to indicate the num- 
ber of yards of fencing 
needed for the, lot in Fig. 18. 

4. In Fig. 19 the whole 
segment is denoted by c. 
Show by measuring on 
squared paper that c = a + b. 

5. In Fig. 19 what rela- 
tion exists between c and 
either a or b? Why? 




FIG. 18 



FIG. 19 



35. Axioms. Exs. 4 and 5, above, illustrate the following 
two axioms : 

VI. The whole is equal to the sum of all its parts. 

VII. The whole is greater than any one of its parts. 



EXERCISES 

1. Draw the segments a = 2.3 cm.; b = 3.2 cm.; c = 1.3cm. 
Draw the sum a + b + c. 

2. Let a, b, and c denote three line segments. Draw a line 
segment to represent 2a + 3b + c ; to represent 4 a -f b -f 2 c ; 
to represent a -\- 3 b -j- 4 c. 



38 GENERAL MATHEMATICS 

3. In Fig. 20, if a, b, and c are three consecutive segments on a 
straight line, such that a = c, show | , I | 

by measuring that a + b = I + c. 

FIG. 20 

4. Show with out measuring that 

a + b = b + c. What axiom does this fact illustrate ? Quote 
the axiom. 

36. Order of terms in addition. The fact that We get the 
same sum when we lay off a segment a and then add b 
that we do when we lay off b first and then add a is a 
geometric illustration of the truth of the commutative law. 
This law asserts that the value of a sum does not change 
when the order of the addends is changed. In the first chap- 
ter we illustrated this principle by a familiar experience 
from arithmetic, as 2 + 5-1-4 = 2 + 4-}- 5. 

EXERCISES 

1. Illustrate the validity of the commutative law by a fact 
from your everyday experience. 

2. Add in the most advantageous way, using the commutative 
law : .376 + 412 + 124 ; 2187 + 469 + 213 ; 36 + 142 + 164. 

37. Difference of two line segments. The difference of 
two line segments may also be found with the compasses. 
To find out how much greater | . 1 

the segment c is than the seg- 
ment b we lay off the segment c 
on a working line (Fig. 21) from 
A to C, then lay the segment A. D b c 

b backward from C toward A. FIG. 21. GEOMETRIC SUBTRAC- 

Then the difference between 

the segments b and c is expressed by the segment AD. In 
equation form this may be written AD = c b. Illustrate 
this method by comparing the lengths of two pencils. 



c-b 



LINEAR MEASUREMENT 39 

EXERCISES 

1. Transfer the line segments of Fig. 21 to squared paper 
and express the difference between c and b in centimeters. 

2. Subtract a line segment 3.5 cm. long from one 6 cm. long. 

3. In Fig. 22 the line segment AB equals the line segment 
MX. Show by measuring that M h N 

AB - MB = MN - MB. \ h^-+ 1 

A B 

4. Ex. 3 is simpler if we ^ IG 22 
write the fact in algebraic 

form, using the small letters. Thus, a + c = b + c. How would 
you show that a = b ? 

5. What axiom of the first chapter is illustrated by Exs. 3 
and 4 ? Quote the axiom. 

6. If a = 3 cm., b = 2 cm., and c 1 cm., construct a line 
segment representing 2 a + 3 b c ; representing 5a 2b + c. 

7. If a, b, and c represent the length of three respective seg- 
ments, construct 3 <z -f- 2 # c ; 4 a + 2 2 c ; 5 a 2 6.+ 3 c. 

8. How long would each of the segments constructed in 
Ex. 7 be if a = 5, b = 4, and c = 3 ? 

38. Coefficient. The arithmetical factor in the term 2# 
is called the coefficient of the literal factor #. When no 
coefficient is written, as in #, we understand the coeffi- 
cient to be 1. Thus, x means 1 x. The coefficient of a 
literal number indicates how many times x 

the literal number is to be used as an ~ ~ 

addend; thus, 5 a: means x+x+x+z+x 
and can be expressed by the equation 5x = x+ x+x-\-x+x. 
Written in this form we see that the use of a coefficient 
is a convenient method of abbreviating. Geometrically a 
coefficient may be interpreted as follows: Let x be the 
length of the segment in Fig. 23. Then the 5 in 5x indi- 
cates that the line segment x is to be laid off five times 



40 GENERAL MATHEMATICS 

consecutively on a working line. 5.r expresses the sum 
obtained by this geometric addition. Find this sum. Usu- 
ally the term "coefficient" means just the arithmetical factor 
in a term, though in a more general sense the coefficient 
of any factor, or number in a term, is the product of all the 
other factors in that term. Thus, in 3 aby the coefficient 
of y is 3 ab, of by is 3 a, of aby is 3. 

EXERCISE 

Give the coefficient in each of the following terms : 3 b ; 
x 7x 8# 9 a-. 



39. Polynomials. An algebraic number consisting of two 
or more terms (each called a monomial), as bx + % y 2 2, 
is a polynomial. The word "polynomial" is derived from 
a phrase which means many termed. A polynomial of two 
terms, as 5 x + 3 #, is a binomial. A polynomial of three 
terms, as 2a-f3J4-4c, is a trinomial. 

EXERCISE 

Classify the following expressions on the basis of the 
number of terms : 

(a) 2 m + 3 n 5 x + r. (c) 6 x + 2 y. 

(b) 6x. (d) a + 2& + 3. 

40. Algebraic addition of similar terms. In simple prob- 
lems we have frequently added similar terms. We shall 
now review the process by means of the following example 

in order to see clearly the law to be used in the more 
complicated additions : 

Add 4:r-!-3a: + 2:r. 

Solution. 4 x can be considered as the sum of four segments each 
x units long. 

Therefore 4x = z + z + r + x. 



LINEAR MEASUREMENT 41 

Similarly, 3 x = x + x + x, 

and 2x = z + x. 

Adding, 

4:X + 3x + 2x=:x + x + x + x + x + x + x + x + x, or 9 z. 
Hence . 4z + 3z + 2z = 9z. 

The preceding example illustrates the law that the sum 
of two or more similar monomials is a monomial whose coeffi- 
cient is the sum of the coefficients of the given monomials and 
which has the same literal factor as the given monomials. 

The advantages of adding numbers according to this 
law may be seen by comparing the two solutions of the 
following problem : 

The tickets for a school entertainment are sold at 250 each 
by a cominittee of five students, A, B, C, D, and E. A reports 
38 sold; B, 42; C, 26; D, 39; and E, 57. At the door 173 
tickets are sold. Find the total receipts. 

Solution I. 



A's receipts, 


38 


x 


$0.25 = 


$9.50 


B's receipts, 


42 


x 


0.25 = 


10.50 


C's receipts, 


26 


X 


0.25 = 


6.50 


D's receipts, 


39 


X 


0.25 = 


9.75 


E's receipts, 


57 


X 


0.25 = 


14.25 


Door receipts, 


173 


X 


0.25 = 


43.25 


Total receipts, 








$93.75 



Since the common factor in the above multiplication is 
25, a simpler solution is obtained if the numbers of tickets 
sold (coefficients) are first added and placed before the 
common factor ; thus, 

Solution II. A's receipts, 38 x $0.25 



B's receipts, 


42 x 


0.25 




C's receipts, 


26 x 


0.25 




D's receipts, 


39 x 


0.25 




E's receipts, 


57 x 


0.25 




Door receipts, 


173 x 


0.25 




Total receipts, 


375 x 


0.25 


= $03.75 



42 GENERAL MATHEMATICS 

EXERCISES 

1. Tickets were sold at c cents ; A sells 12; B, 15; C, 36; 
and D, 14. There were 112 tickets sold at the gate. Find the 
total receipts. 

2. Express as one term 3-7 + 5-7-f4-7. 

NOTE. A dot placed between two numbers and halfway up in- 
dicates multiplication and is read "times." Do not confuse with 
the decimal point. 

3. Can you add 3 7 + 14 2 -f 5 4 by the short cut above? 

4. Indicate which of the following sums can be written in 
the form of monomials : 3x + 5a;; x + 7 x + 3; 13 + 5 -|- 3 ; 
3 + 2 1 + 4. 

5. Add as indicated : (a) 3 x + 20 x + 17 x + 9 x + ~x + 3 x ; 
(b) 3y + y + 15y+ily + 2y; (c) 9* + 3s + 3s + 4 .s -j- 2s ; 



6. The school's running track is / feet. While training, a 
boy runs around it five times on Monday, six on Tuesday, ten 
on Wednesday, seven 011 Thursday, six on Friday, and nine on 
Saturday morning. How many feet does he run during the week ? 

41. Subtraction of similar monomials. The law in sub- 
traction is similar to the law in addition and may be 
illustrated as follows: 
Subtract 2 x from 5 x. 
Solution. 5x = x + x + x + x + x. 

2 x = x + x. 
Subtracting equal numbers from equal numbers, 

5 x 2 x = x + x + x, or 3 x. 
Hence 5x 2x = 3 x. 

The preceding example illustrates the law that tJie differ- 
ence of two similar monomials is a monomial having a coefficient 
equal to the difference of the coefficients of the given monomials 
and having the same literal factor. 



LINEAR MEASUREMENT 



43 




EXERCISES 

1. Subtract 3 b from 146. 

2. Write the differences of the following pairs of numbers as 
monomials: lOce 3ic; 13x 5x; 12z 3z; VI k 5k; 2.68 r 
-0.27/-; 1.03a-0.08a; fe-Je; f* i*. 

3. The following exercises require both addition and sub- 
traction. Write each result as a single term : <ix-}-6x 2x; 
13x 2x + 3x; 11.5c + 2.3c - c; Ja + f a J a. 

42. Triangle; perimeter. If three points, as A, B, and 
(7 (Fig. 24), are connected by line segments, the figure 
formed* is a triangle. The 
three points are called ver- 
tices (corners) of the triangle, 
and the three sides a, 6, and A 
c are the sides of the -tri- 
angle. The sum of the three sides, as a -f- b + c (the 
distance around), is the perimeter of the triangle. 

EXERCISES 

1. A yard has the form of an equal-sided (equilateral) 
triangle, each side being x rods long. How many rods of 
fence will be needed to inclose it ? 

2. What is the sum of the sides 
of a triangle (Fig. 25) whose sides 
are 2 x feet, 2 x feet, and 3 x feet 
long? Express the sum as a cer- 
tain number of "times x. 

3. What is the sum of the three sides 3b, 4b, and 6b of a 
triangle ? Express the result as one term. 

4. What is the perimeter of a triangle whose sides are 2x, 
8 x } and 9 x ? Let p be the perimeter ; then write your answer 
to the preceding question in the form of an equation. 




2x 



3x 
FIG. 25 



44 



GENERAL MATHEMATICS 



43. Polygons. A figure, as ABODE (see Fig. 26), formed 
by connecting points, as A, J5, C, D, and E, by line 
segments, is a polygon. 
The Greek phrase from 
which the word "polygon" 
is derived means many cor- 
nered. Polygons having 3, 
4, 5, 6, 8, 10, ., n sides 
are called triangle, quadri- 
lateral, pentagon, hexagon, 
octagon, decagon, - - ., n-gon FlG 2 6. A POLYGON 

respectively. The sum of 

the sides of a polygon is its perimeter. When all the 
sides of a polygon are equal it is said to be equilateral. 




EXERCISES. 



1. What is the perimeter of each of the polygons in Fig. 27? 
In each case express the result in the form of an equation ; 
thus, for the first quadrilateral p = 12 x. 




IX 



FIG. 27 



2. Show by equations the perimeter p of the polygons in 
Fig- 28. 

3. Show by equations the 
perimeter of an equilateral 
quadrilateral whose side is 
11; 9; s; x; b; z\ 2e; 
9 + 3; a + 5; a + d; x + 7 : 
x-fy. 



10 ,2 






c 






d 




FIG. 28 



LINEAR MEASUREMENT 45 

4. Name the different figures whose perimeters might be 
expressed by the following equations : 

p = 3 s, p = 5 s, j) = 7 s, p = 9 s, p = 12 s, p = 20 s, 
p = 4 s, p = 6 s, p = 8 s, p = 10 s, p = 15 s, p = ns. 

5. Assume that all the figures in Ex. 4 are equilateral. 
Find out how many of your classmates can give the name of 
each polygon. 

6. Assume that at least six of the polygons in Ex. 4 are 
not equilateral. Sketch the figures of which the given equa- 
tions may be the perimeters. 

7. What is the perimeter of each figure of Ex. 4 if s = 2 in. ? 
ifs = 3cm.? if s = 4yd.? ifs = 5ft.? 

8. Determine the value of s in the equations p = 3 s, p = 4 s, 
p = 5s, p = 6s, p = 10s, and p = 15s if in each case the 
perimeter is 120 in. 

9. What is the side of an equilateral hexagon made with a 
string 144 in. long ? (Use all the string.) 

10. Show by sketches polygons whose perimeters are ex- 
pressed by p = 8 a + 6 ; by^? = 4 + 12; by^? = 66 + 6a; 
by^> 4a + 25; by p = 3a -\- 2b. 

11. Find the value of the perimeters in Ex.10 if a = 3 
and b = 2 ; if a = 5 and b = 3 ; if a = 1 and b = 5. 

12. If x = 2 and y 3, find the value of the following 
expressions : 3x + ?/ ; 3x y; 3x 2 ?/ ; 2x 3|- ; 4 a; 2-^ ?/ ; 
2.25 x-y; 2.27 x- 1.12 y. 

SUMMARY 

44. This chapter has taught the meaning of the follow- 
ing words and phrases : line segment, point, measurement 
of length, unit segment, standard unit, ratio, metric system, 
coincide, intersect, equal segments, unequal segments, com- 
mutative law, coefficient, polynomial, binomial, trinomial, 



46 GEKEKAL MATHEMATICS 

triangle, vertex of a polygon, vertices of a polygon, perim- 
eter, sides of a triangle, polygon, quadrilateral, pentagon, 
hexagon, octagon, decagon, n-gon, equilateral. 

45. Axioms. The following axioms were illustrated: 

VI. The whole is equal to the sum of all its parts. 

VII. The whole is greater than any of its parts. 

46. The following instruments have been used in measur- 
ing line segments : the ruler, the compasses, and squared paper. 

47. The following symbols were used :. = meaning does 
not equal; < meaning is less than; > meaning is greater 
than ; and a dot, as in 3 5, meaning times, or multiplied by. 

48. A point is determined by two intersecting lines. 

49. The metric system has certain advantages over our 
English system. 

50. The practical difficulty of precise measuring has 
been pointed out. Five possible errors were enumerated. 
Measurement implies the determination of a ratio. 

51. The sum of two segments was found with the com- 
passes. A law was discovered to serve as a short cut in 
algebraic addition. 

52. The difference of two segments was found with the 
compasses and the law for algebraic subtraction was stated. 

53. The Addition and Subtraction laws of Chapter I 
were illustrated geometrically. 

54. The perimeter of a figure may be expressed by an 
equation. 

55. The chapter has taught how to find the value of an 
algebraic number when the value of the unknowns are 
given for a particular cage; for example, how to find the 
value of 3 x + 2 y when x- == 1 and #==2. 



CHAPTER III 

PROPERTIES OF ANGLES 

56. Angle. If a straight line, as OX in either of the 
drawings of Fig. 29, rotates in a plane about a fixed point, 
as 0, in the direction indicated by the arrowheads (counter- 
clockwise) until it reaches the position OT, it is said to 



x 




FIG. 29. ILLUSTRATING THE DEFINITION OF AN ANGLE 



turn through the angle XOT. Thus, an angle is the amount 
of turning made by a line rotating about a fixed point in a 
plane (flat surface}. Note that as the rotation continues, 
the angle increases. 

57. Vertex. The fixed point (Fig. 29) is called the 
vertex of the angle. (The plural of "vertex" is "vertices.") 

47 



48 GENERAL MATHEMATICS 

58. Initial side ; terminal side. The line OX (Fig. 29) 
is called the initial side of the angle. The line OT is called 
the terminal side of the angle. 

59. Symbols for " angle." The symbol for "angle" is Z; 
for " angles," A Thus, " angle XOT" is written /^XOT. 

60. Size of angles. From the definition of an angle given 
in Art. 56 we see that it is possible for the line OX to stop 
rotating (Fig. 29) so that the angle may contain any given 
amount of rotation (turning). 

EXERCISE 

Draw freehand an angle made by a line OX which has 
rotated one fourth of a complete turn ; one half of a complete 
turn ; three fourths of a complete turn ; one complete turn ; 
one and one-fourth complete turns. 

61. Right angle ; straight angle ; perigon. If a line 
rotates about a fixed point in a plane so as to make 

T one fourth of a complete turn, the angle formed 
is called -a right angle (rt. Z) (see Fig. 30, (a)). 



-X 




O 
(a) Right Angle (b) Straight Angle (c) Perigon 

FIG. 30. THREE SPECIAL ANGLES 

If the line makes one half of a complete turn, the angle 
formed is called a straight angle (st. Z) (see Fig. 30, (b)) ; if 
the line makes a complete turn, the angle formed is called 
& perigon (see Fig. 30, (c)). 

EXERCISES 

1. Draw freehand an angle equal to 1 right angle; 2 right 
angles ; 3 right angles ; 4 right angles. 

2. Draw freehand an angle equal to 1 straight angle; 
2 straight angles ; 1|- straight angles ; 2^ straight angles. 



PKOPEKTIES OF ANGLES 



49 



3. How many right angles are there in a half turn of a 
rotating line ? in a whole turn ? in 5 turns ? in 3^ turns ? ' 

in x turns ? in - turns ? in - turns ? 
2 4 

4. How many straight angles are there in a half turn of a 
rotating line ? in a whole turn ? in 1^ turns ? in 1| turns ? 

5x 

in x turns ? in 2x turns ? in - turns ? 

4 

5. Through how many right angles does the minute hand of 
a watch turn in 3 hr. ? in 3^ hr. ? in 4 hr. ? in 1\ hr. ? in 
x hours ? in 15 min. ? in 30 min. ? in 5 min. ? in 10 min. ? 

6. How many straight angles are there in a perigon ? in 
a right angle ? in 10 right angles ? in y right angles ? 

62. Acute angle ; obtuse angle ; reflex angle. Angles 
are further classified upon the basis of their relation to 

T 





FIG. 31. ACUTE ANGLE 



FIG. 32. OBTUSE ANGLE 



the right angle,' the straight angle, and the perigon. An 

angle less than a right angle is called an acute angle, 

Fig. 31. An angle 

which is greater 

than a right angle 

and is less than- 'H 

a straight angle 

is called an obtuse 

angle, Fig. 32. An 

angle greater than 

a straight angle 

and less than a perigon is called a reflex angle, Fig. 33. 




FIG. 33. REFLEX ANGLE 



GENERAL MATHEMATICS 



EXERCISES 

4 

1. Draw an acute angle ; an obtuse angle ; a reflex angle. 

2. Point out, in the classroom, examples of right angles ; 
of obtuse angles. 

C\ 
C 

D 




(a) (b) (c) 

FIG. 34. ILLUSTRATING THE VARIOUS KINDS OF ANGLES 

3. In the drawings of Fig. 34 determine the number of acute 
angles ; of right angles ; of obtuse angles ; of reflex angles. 

63. Notation for reading angles. There are three common 
methods by which one may denote angles: (1) Designate 
the angle formed by two lines OX 
and OT as the " angle XOT" or the 
" angle TOX" (Fig. 35). Here the 
first and last letters denote points 
on the lines forming the angle, and 
the middle letter denotes the point 
of intersection (the vertex). In read- 
ing " angle XOT" we regard OX as 
the initial side and OT as the terminal side. (2) Denote the 
angle by a small letter placed as # in Fig. 36. In writing 
equations this method is the B 

most convenient. (3) Denote 
the angle by the letter which 
is written at the point of in- 
tersection of the two sides of 
the angle, as " angle A" (Fig. 36). This last method is used 
only when there is no doubt as to what angle is meant. 



FIG. 35 



FIG. 36 



PKOPERTIES OF ANGLES 

EXERCISE 



51 



In the drawings of Fig. 37, below, select three angles and 
illustrate the three methods of notation described above. 




(b) 




FIG. 37 



64. Circle. If a line OX be taken as the initial side 
of an angle (see Fig. 38) and the line be rotated one 
complete turn (a peri- 
gon), any point, as P, 
on the line OX will 
trace a curved line 
which we call a circle. 
Thus, a circle is a 
closed curve, all points 
of which lie in the 
same plane and are 
equally distant from a 
fixed point. FIG. 38. THE CIRCLE 




52 GENERAL MATHEMATICS 

65. Center ; circumference. The fixed point is the 
center of the circle. The length of the curve (circle) is 
called the circumference (distance around) of the circle. 

66. Radius ; diameter. A line drawn from the center 
of a circle to any point on the circle is a radius. Thus, 
OP is a radius of the circle in Fig. 38. A line connecting 
two points on the circle and passing through the center of 
the circle is called a diameter. 

From the definition of " radius " given above it is clear 
that in a given circle or in equal circles one radius has 
the same length as any other. Thus we obtain the follow- 
ing important geometric relation, Radii of the same circle 
or of equal circles are equal. ("Radii" is the plural of 
" radius.") 

67. Arc; to intercept; central angle. An arc is a part 
of a circle. If two radii are drawn from the center of the 
circle to two different points on the circle, they cut off an 
arc on the circle. The symbol for." arc " is ""^ Thus, AB is 
read " the arc AB" The angle formed at the center of the 
circle is said to intercept the arc. The angle at the center 
is called a central angle. 

68. Quadrant; semicircle. An arc equal to one fourth 
of a circle is called a quadrant. An arc equal to one half 
of a circle is called a semicircle. 

EXERCISES 

1 . How does a diameter compare in length with a radius ? 

2. How many quadrants in a semicircle ? in a circle ? In 
what connection have we mentioned the idea expressed by the 
word " quadrant " ? 



PROPERTIES OF ANGLES 



53 



69. Degrees of latitude and longitude. The use that is 
made of the circle in geography is no doubt familiar to 
all of us. The prime meridian, that passes through Green- 
wich, England (see Fig. 39), is the zero meridian. We 
speak of places lying to the west of Greenwich as being 
in west longitude and of those lying to the east of 
Greenwich as being in east longitude (see Fig. 39). Since 
it takes the earth twenty-four hours to make one complete 
rotation, the sun apparently passes over one twenty-fourth 



N.P. 



N.P. 




S.P. 
LATITUDE AND LONGITUDE 

of the entire distance around the earth every hour. Thus, 
points lying a distance of one twenty -fourth of a complete 
turn apart differ in time by an hour. 

In order to carry the computations further the entire 
circle is divided into three hundred and sixty equal parts, 
each of which is called a degree (1) of longitude. In 
order to express fractional parts of the unit each degree 
is divided into sixty minutes (60') and each minute into 
sixty seconds (60"). With this agreement the longitude 
of a place is determined. 

The position of a place north or south of the equator is in- 
dicated by the number of degrees of north or south latitude. 



54 GENERAL MATHEMATICS 

EXERCISES 

1. What is the greatest longitude a place can have ? the 
greatest latitude ? 

2. How many seconds are there in a degree of longitude ? 
in a degree of latitude ? * 
( $7 What is the length of a degree arc on the earth's 
surface ? of a minute arc ? of a second arc ? Try to find out 
how accurately the officers of a ship know the location of 
the ship out in mid-ocean. 

4. How would you read 25 14' west longitude ? 33 5' 17" 
north latitude ? 

5. Compare the method of locating by latitude and longitude 
to the method used in locating a house in a large city. 

6. Find out in what longitude you live ? in what latitude ? 

70. Amount of rotation determines the size of an angle. 
If we remember that an angle is formed by rotating a line 
in a plane about a fixed point, it will be clear that the 




Fio. 40 

size of the angle depends only on the amount of turning, 
not upon the length of the sides. Since the sides may 
be extended indefinitely, an angle may -have short or long 
sides. In Fig. 40 the angle A is greater than angle B, but 
the sides of angle B are longer than the sides of angle A. 

71. Measurement of angles; the protractor. In many 
instances the process of measuring angles is as important 
as that of measuring distances. An angle is measured 
when we find how many times it contains another angle 
selected as a unit of measure. 



PROPERTIES OF ANGLES 



55 




FIG. 41. THE PROTRACTOR 



The protractor (Fig. 41) is an instrument devised for 
measuring and constructing angles. The protractor com- 
monly consists of a semicircle divided into one hundred 
and eighty equal parts. Each of these equal parts is called 
a degree of arc (1). In the geography work referred to in 
Art. 69, the unit for longitude and latitude was the degree 
of arc. In the measure- 
ment of angles we shall 
consider a unit corre- 
sponding to a unit of 
arc and called a degree 
of angle. 

If straight lines are 
drawn from each of 
these points of division 
on the semicircle to the 
center 0, one hundred 
and eighty equal angles are formed, each of which is a 
degree of angle (I )-. Thus, the unit of angular measure 
is the degree. A degree is divided into sixty equal parts, 
each of which is called a minute (!'). 

Each minute is divided into sixty equal parts, eack 
of which is called a second (1"). Of course the minute 
and the second graduations are not shown on the pro- 
tractor. Why not ? 

EXERCISES 

1 . How many degrees in a right angle ? in a straight angle ? 
in a perigon ? 

2. A degree is what part of a right angle ? of a straight 
angle ? of a perigon ? 

3. What angle is formed by the hands of a clock at 3 o'clock ? 
at 6 o'clock ? at 9 o'clock ? at 12 o'clock ? at 4 o'clock ? at 
7 o'clock ? at 11 o'clock ? 



56 GENERAL MATHEMATICS 

4. Give a time of day when the hands of a clock form a 
right angle; a straight angle. 

5. What is the correct way to read the following angles : 
15 17' 2"? 5 0' 10"? 

6. How many degrees are there in three right angles ? in 
four straight angles ? in one third of a right angle ? in two 
thirds of a straight angle ? in one fifth of a right angle ? in 
x straight angles ? in y right angles ? in 2 x right angles ? 

7. Ordinary scales for weighing small objects are sometimes 
made with a circular face like a clock face. The divisions of 
the scale indicate pounds. If the entire face represents 12 lb., 
what is the angle between two successive pound marks ? 

8. What is the angle between two successive ounce marks 
on the face of the scale in Ex. 7 ? 

72. Measuring angles ; drawing angles. The protractor 
may be used to measure a given angle. Thus, to measure 
a given angle x place the protractor so that the center of 
the protractor (point in Fig. 42) falls upon the vertex 
and make the straight edge of the protractor coincide 
with (fall upon) the initial side of the given angle x. Now, 
observe where the terminal side of the given angle intersects 
(crosses) the rim of the protractor. Read the number of 
degrees in the angle from the scale on the protractor. 

EXERCISES 

1. Draw three different angles, one acute, one obtuse, and 
one reflex. Before measuring, estimate the number of degrees in 
each angle. Find the number of degrees in each angle by means 
of the protractor. Compare the results with the estimates. 

2. Draw freehand an angle of 30; of 45; of 60; of 90; 
of 180 ; of 204. Test the accuracy of the first four angles by 
means of the protractor. 



PEOPERTIES OF ANGLES 



57 




The protractor is also useful in constructing angles 
of a required size ; for example, to construct an angle 
of 42 draw 
a straight line 
OX (Fig. 42) 
and place the 
straight edge 
of the pro- 
tractor on the 

FIG. 42. MEASURING ANGLES WITH A PROTRACTOR 
that the center 

rests at 0. Count 42 from the point on OX where the 
curved edge touches OX and mark the point A. Connect 
A and 0, and the angle thus formed will contain 42. 

EXERCISES 

1. Construct an angle of 25; of 37; of 95; of 68; of 
112 ; of 170. Continue this exercise until you are convinced 
that you can draw any required angle. 

2. Construct an angle equal to a given angle ABC by means 
of the protractor. 

HINT. Draw a working line MN. Measure the angle ABC. 
Choose a point P on the line MN as a vertex and construct an 
angle at P containing the same number of degrees as the given 
angle ABC. The angle is then constructed as required. 

3. Draw a triangle. How many angles does it contain ? 
Measure each with the protractor. 

73. How to measure angles out of doors. It is possible 
to measure angles out of doors by means of a simple field 
(out-of-door) protractor, so that some simple problems in sur- 
veying can be solved. Such a field protractor may be made 
by a member of the class, as shown on the following page. 



58 



GENERAL MATHEMATICS 



Secure as large a protractor as possible and fasten it on 
an ordinary drawing board. Attach the board to a camera 
tripod (if this is not to be had, a rough tripod can be 
made). Make a slender pointer which may be attached at 
the center of the circle with a pin so that it may swing freely 
about the pin as a pivot. Place two inexpensive carpenter's 
levels on the board, and the instrument is ready for use. 

Thus, to measure an angle ABC (suppose it to be an 
angle formed by the intersection of an avenue, BA, and a 
street, -BC), first put the board in a horizontal position 
(make it stand level). Then place the center of the circle 
over the vertex of ; 

the angle to be meas- 
ured and sight in 
the direction of each 
side of the angle, 
noting carefully the 
reading on the pro- 
tractor. The number 
of degrees through 
which the pointer 
is turned in passing 
from the position of 
BA to that of BC 
is the measure of 
angle ABC. 

74. Transit. When 
it is important to 
secure a higher de- 
gree of accuracy than is possible with the instrument 
described in Art. 73, we use an instrument called a transit 
(Fig. 43). This instrument is necessary in surveying. 
Three essential parts of the transit are (1) a horizontal 




FIG. 43. THE TRANSIT 



59 

graduated circle for measuring angles in the horizontal 
plane (see D in Fig. 43) ; (2) a graduated circle, C, for 
measuring angles in the vertical (up-and-down) plane ; and 
(3) a telescope, AB, for sighting in the direction of the 
sides of the angle. For a fuller description of the transit 
see a textbook in trigonometry or surveying. 

HISTORICAL NOTE. The division of the circle into three hundred 
and sixty degrees and each degree into sixty minutes and each 
minute into sixty seconds is due to the Babylonians. Cajori cites 
Cantor and others somewhat as follows : At first the Babylonians 
reckoned the year as three hundred and sixty days. This led them 
to divide the circle into three hundred and sixty degrees, each degree 
representing the daily part of the supposed yearly revolution of the 
sun around the earth. Probably they were familiar with the fact that 
the radius could be applied to the circle exactly six times and that 
as a result each arc cut off contained sixty degrees, and in this way 
the division into sixty equal parts may have been suggested. The 
division of the degree into sixty equal parts called minutes may have 
been the natural result of a necessity for greater precision. Thus the 
sexagesimal system may have originated. " The Babylonian sign * is 
believed to be associated with the division of the circle into six equal 
parts," and that this division was known to the Babylonians seems 
certain " from the inspection of the six spokes in the wheel of a royal 
carriage represented in a drawing found in the remains of Nineveh." 

Henry Briggs attempted to reform the system by dividing the 
degree into one hundred minutes instead of into sixty, and although 
the inventors of the metric system are said to have proposed the 
division of the right angle into one hundred equal parts and to 
subdivide decimally, instead of the division into ninety parts, we 
have actually clung to the old system. However, there is a tend- 
ency among writers to divide each minute decimally ; for example, 
52 10.2' instead of 52 10' 12". See Cajori, " History of Elementary 
Mathematics," 1917 Edition, pp. 10, 43, and 163. 

75. Comparison of angles. In order to make a comparison 
between two angles, we place one over the other so that the 
vertex and the initial side of one coincide with the vertex and 
the initial side of the other. If the terminal sides coincide, 



60 



GENERAL MATHEMATICS 



the angles are equal ; if the terminal sides do not coincide, 
the angles are unequal assuming, of course, in both cases, 
that each of the two angles compared is less than 360. In 
the exercises and articles that follow we consider no angle 
greater than 360. 

EXERCISES 

1. Compare angles x, y, and 2, in Fig. 44, and arrange them 
in order as to size. 

HIXT. Make a tracing of each on thin paper and try to fit each 
on the other. 




FIG. 44 

2. Construct an angle equal to a given angle ABC. Lay a thin 
sheet of paper over the angle ABC and make a tracing of it. Cut 
out the tracing and paste it to another part of the paper. The 
angle thus shown is equal to the angle ABC. 

3. Try to draw freehand two equal angles. 
Test your drawings by the method of Ex. 1. 

4. Draw freehand one angle twice as 
large as another. Test your drawings with 
the protractor. 




FIG. 45 



76. Adjacent angles ; exterior sides. 
Angles x and y in Fig. 45 are two angles 
which have a common vertex and a com- 
mon side between them. The angles x and y are said to be 
adjacent angles. Thus, adjacent angles are angles that have 
the same vertex and have a common side between them. 
The sides OT and OR are called the exterior sides. 



PKOPERTIES OF ANGLES 



61 



E 



EXERCISES 

1. Indicate which angles in Fig. 46 are adjacent. Point out 
the common vertex and the common side in each pair of 
adjacent angles. 

2. Draw an angle of 45 adjacent 
to an angle of 45; an angle of 30 
adjacent to an angle of 150; an 
angle of 35 adjacent to an angle 
of 80. 

3. Do you notice anything partic- 
ularly significant in any of the parts 
of Ex. 2 ? 

4. Draw an angle of 30 adjacent 
to an angle of 60. What seems to 
be the relation between their exterior 
sides ? Does this relation need to 
exist in order that the angles shall 

be adjacent ? What total amount of turning is represented ? 

77. Geometric addition and subtraction of angles. Exs. 2 

and 4, above, suggest a method for adding any two given 
angles. Thus, to add a given angle y to a given angle x, 

B 




FIG. 46 





FIG. 47. GEOMETRIC ADDITION OF ANGLES 

Fig. 47, angle y is placed adjacent to angle z, and the re- 
sulting angle is called the sum of x and y. The angles may 
be transferred to the new position either by means of tracing 
paper or, more conveniently, by means of the protractor. 



62 GENERAL MATHEMATICS 

EXERCISE 

Add two angles by placing them adjacent to each other. 

We may also find the difference between two angles. 
In Fig. 48 the two given angles are x and y. Place tin- 
smaller angle, y, on the larger, a:, so that the vertices and 




o 




FIG. 48. GEOMETRIC SUBTRACTION OF ANGLES 

one pair of sides coincide. The part remaining between the 
other two sides of x and y will be the difference between 
x and y. Thus, in Fig. 48 we obtain Zz /.y=/.AOC. 

EXERCISES 

1. Draw three unequal angles x, y, and 2, so that y~>x and 
x > z. Draw an angle equal to x + y + z ; equal to y x + z ; 
equal to y -\- x - z. 

2. Draw an angle of 60 and draw another of 20 adjacent 
to it. What is their sum ? Fold the 20-degree angle over the 
60-degree angle (subtraction) and call the difference x. What 
is the equation which gives the value of x ? 

78. Construction problem. At a given point on a given 
line to construct by means of ruler and compasses an 
angle equal to a given angle. In this construction we 
make use of the following simple geometric relation be- 
tween central angles and their intercepted arcs : In the 
same circle or in equal circles equal central angles intercept 



PROPERTIES OF ANGLES 63 

equal arcs on the circle. For example, if the central angle 
contains nineteen angle degrees, then the intercepted arc 
contains nineteen arc degrees. 

The student may possibly see that this geometric rela- 
tion is implied in our definitions of Art. 71. However, 
the two following paragraphs will assist him in under- 
standing its application. 

Make a tracing of the circle and the angle ABC .in 
Fig. 49, (a), and place B upon E in Fig. 49, (b). The 
angles must coincide because they are given equal. Then 
the circle whose center 
is B (circle B) must 
coincide with the cir- 
cle whose center is E 
(circle -E 1 ), because the 
radii of equal circles 
are equal. Then A will 

FIG. 49 
fall on Z>, and C on F; 

that is, the arc CA will fall on the arc FD, and the arcs 
are therefore equal. 

It is easy to see that the following statement is also 
true : In the same circle or in equal circles equal arcs on the 
circle are intercepted by equal central angles. For circle M 
can be placed on circle E so that arc CA coincides with 
arc FD, since these arcs are given equal, and ''so .that B 
falls on E. A will fall on Z>, and C on F. Then the angles 
must. coincide and are therefore equal. 

The two preceding geometric relations make clear why 
the protractor may be used to measure angles as we did 
in Art. 71. The method used there is based upon the idea 
that every central angle of one degree intercepts an arc 
of one degree on the rim of the protractor ; that is, when 
we know the number of degrees in an angle at the center 




64 



GENEKAL MATHEMATICS 



of a circle we know the number of degrees in the arc 
intercepted by its sides, and vice versa. 

The idea can be expressed thus: A central angle is 
measured by the arc intercepted by its sides (when angular 
degrees and arc degrees are used as the respective units 
of measure). 

How many degrees of an arc are intercepted by a central 
angle of 30 ? of 40 ? of 60.5 ? of n ? 

We are now ready to proceed with our problem: At 
a given point on a given line to construct by means of 
ruler and compasses an angle equal to a given angle. 




FIG. 60. CONSTRUCTING AN ANGLE EQUAL TO A GIVEN ANGLE 



Construction. Let DEF in Fig. 50 be the given angle and let P 
be the given point on the given line AB. 

With E as a center and ER as a radius draw a circle. With P as 
a center and with the same radius {ER^ draw another circle. Place 
the sharp point of the compasses at R and cut an arc through M. 
With S as a center and the same radius cut an arc at N. 

The Z.BPC is the required angle. Why? 



EXERCISES 



1. Check the correctness of your construction for the pre- 
ceding directions by measuring with a protractor. 

2. How many ways have we for constructing an angle equal 
to a given angle ? 



PROPERTIES OF ANGLES 65 

3. Construct two angles, designating one of them as contain- 
ing a degrees and the other as containing b degrees ; using the 
angles a and b, construct an angle containing a + b degrees ; 
construct an angle containing 2 a -f- b degrees. 

4. Choose a and b in Ex. 3 so that a >b, then construct an 
angle equal to the difference of the two given angles. 

5. Construct an angle equal to the sum of three given angles. 

79. Perpendicular. We have seen in Art. 61 how right 
angles are formed by rotation. If two lines form right 
angles with each other, they are said to be perpendicular 
to each other. The symbol 

for " perpendicular " is _L. ^ 

80. Construction problem. 
At a given point on a given 
line to erect a perpendic- 
ular to that line by using A 

ruler and compasses. c \ I D 

FlG. 51. HOW TO ERECT A 

Construction. Let AB be the PERPENDICULAR 

given line and P the given 

point (Fig. 51). With P as a center and with a convenient radius 
draw arcs intersecting AB~&t C and D. 

With C and D as centers and with a radius greater than \ CD 
draw arcs intersecting at E. Draw EP. Then EP^s the required 
perpendicular. 

EXERCISES 

1. Test the accuracy of your construction in the problem of 
Art. 80 by using a protractor. 

2. Why must the radius CE in Fig. 51 be greater than ^ CD? 

3. In Fig. 52 how < =- 

A 
would you draw a ^ IG 2 

perpendicular to the 

line AB at the point A? Do the construction work on paper. 



06 GENERAL MATHEMATICS 

81. Construction problem. How to bisect a given line 
segment AB. 

Construction. Let AB be the given 
line segment (Fig. 53). With A as 
a center and with a radius greater 

A I I 171 

than ^ AB describe arcs above and * 
below AB. With B as a center and 
with the same radius as before de- 
scribe arcs above and below and \ /~ 
intersecting the first arcs at C and D. 
Draw CD. Then E is the point of FIG. 53. How TO BISECT 
bisection for AB. * LINE SEGMENT 

82. Perpendicular bisector. The line CD in. Fig. 53 is 
called the perpendicular bisector of AB. 

EXERCISES 

1. How may a line be divided into four equal parts ? into 
eight equal parts ? 

2. Draw a triangle all of whose angles are acute (acute- 
angled triangle). Construct the perpendicular bisectors of 
each of the three sides of the triangle. 

3. Cut out a paper triangle and fold it so as to bisect 
each side. 

4. Draw a trianglfe in which one angle is obtuse (obtuse- 
angled triangle) and draw the perpendicular bisectors of the 
three sides. 

5. Draw a triangle in which one angle is a right angle and 
construct the perpendicular bisectors of the sides. 

6. Draw a triangle ABC. Bisect each side and connect each 
point of bisection with the opposite vertex. 

83. Median. A line joining the vertex of a triangle to 
the mid-point of the opposite side is called a median. 



PKOPEKTIES OF ANGLES 



EXERCISE 
Draw a triangle ; construct its medians. 

84. Construction problem. From a given point out- 
side a given line to drop a perpendicular to that line. 

Construction. Let AB be the given line and P the given point 
(Fig. 54). With P as a center and p 

with a radius greater than the dis- 
tance from P to AB describe an arc 
cutting AB at M and R. With M 
and R as centers and with a radius 
greater than \ MR describe arcs 
either above or below (preferably be- 
low) the line AB. Connect the point 
of intersection E with P. Then the 
line PD is perpendicular to AB, as 
required. Test the accuracy of your 
work by measuring an angle at D. 




\/E 



FIG. 54. How TO DROP A. 
PERPENDICULAR 



EXERCISES 

1 . Why is it preferable to describe the arcs in Fig. 54 below 
the line AB ? 

2. Draw a triangle ABC all of whose angles are acute and 
draw perpendiculars from each vertex to the opposite sides. 

85. Altitude. An altitude of a triangle is a line drawn 
from a vertex perpendicular to the opposite side. 



EXERCISES 

1. Draw a triangle in which one angle is obtuse and draw 
the three altitudes. 

2. Draw a triangle in which one angle is a right angle and 
draw the three altitudes. 

3. When d.o the altitudes fall inside a triangle ? outside? 



68 



GENERAL MATHEMATICS 



86. To bisect a given angle. Suppose angle ABC to 
be the given angle (Fig. 55). With the vertex B as a 
center and with a convenient radius draw an arc cutting 
HA and BC at X and Y re- 
spectively. With X and 1' 
as centers and with a radius 
greater than ^ XY draw arcs 
meeting at D. Join B and D. 
Then BD is the bisector of 
/.ABC. 




FIG. 55. How TO BISECT 
AN ANGLK 



EXERCISES 

1. Bisect an angle and check by folding the paper so that 
the crease will bisect the angle. 

2. Bisect an angle of 30 ; of 45 ; of 60 ; of 5)0. 

3. Divide a given angle into four equal parts. 

4. Draw a triangle whose angles are all acute and bisect 
each of the angles. 

5. Draw a triangle in which one angle is obtuse and bisect 
each of the angles ; do the same for a triangle in which one 
angle is a right angle. 

87. Parallel lines. AB and CD in Fig. 5(3 have had 
the same amount of angular rotation from the initial line 
EF. Thus, they have the 
same direction and are said 
to be parallel. The symbol 
for ''parallel" is II. Thus, 
AB II CD is read "AB is par- 
allel to CD" 




A C 

FIG. 56. PARALLEL LINES 



88. Corresponding angles ; 
transversal. Angles x and 
// in Fig. 56 are called corresponding angles. The line EF 




69 

is called a transversal. It is clear that the lines are par- 
allel only when the corresponding angles are equal and 
that the corresponding angles are equal only when the 
lines are parallel. 

EXERCISES 

1. Draw figures to illustrate the importance of the last 
statement in Art. 88, above. 

2. Point out the parallel lines you can find in the classroom. 

89. Construction problem. How to draw a line parallel 
to a given line. 

Construction. Choose a point P 
outside the given line A B in Fig. 57. 

Draw a line through P so as to form 

a convenient angle x with AR. Call 

the point of intersection D. At P, 

using DP as initial line, construct an 

angle y equal to angle x (as shown) 

by the method of Art. 78. Then PR j[ 

and AB are parallel because they 

have had the same amount of rota- FIG. 57. How TO DKAW 

tion from the initial line PD. PARALLEL LISKS 

EXERCISES 

1. Construct a line parallel to a given line through a given 
point outside the line. 

2. A carpenter wants a straight-edge board to have parallel 
ends. He makes a mark across each end with his square. Why 
will the ends be parallel ? 

3. In Fig. 57 if angle x = 60, what is the number of degrees 
in Z. y ? Give a reason for your answer. 

4. Two parallel lines are cut by a transversal so as to form 
two corresponding angles (x + 125) and (3 x + 50). Find x and 
the size of each angle. Make a drawing to illustrate your work. 




GENERAL MATHEMATICS 




FIG. 58 



5. In Fig. 58 if AB II CD, what other angles besides x and y 
are equal corresponding angles ? 

6. In Fig. 58, /.x = Z.y. Bisect 
Z x and Z y and show that these 
bisectors are parallel to each other. 

90. Parallelogram. If onepair 
of parallel lines cross (intersect) 
another pair, the four-sided 
figure thus formed is called a 
parallelogram ; that is, a paral- 
lelogram is a quadrilateral whose opposite sides are parallel. 

91. How to construct a parallelogram. If we remember 
the method used in Art. 89 for constructing one line parallel 
to another, it will be 

easy to construct a 
parallelogram. Thus, 
draw a working line 
AB (Fig. 59). Draw 
AR making a conven- 
ient angle with AB. 
Through any point, as 
P, on AR draw a line 
P V parallel to AB. Through any point M on A B draw a line 
MT parallel to AR. The figure AMSP is a parallelogram, 
for its opposite sides are parallel. 

92. Rectangle. If one of the in- 
terior angles of a parallelogram is a 
right angle, the figure is a rectangle 
(Fig. 60). Thus, a rectangle is a 

parallelogram in which one interior angle is a right angle. 





FIG. 59. How TO CONSTRUCT A 
PARALLELOGRAM 



FIG. 60 




PROPERTIES OF ANGLES 71 

EXERCISE 

Show that all the angles of a rectangle are right angles. 
HINT. Extend the sides of the rectangle. 

93. Square. If all the sides of a rec- 
tangle are equal, the figure is called a 

square (Fig. 61). 

FIG. 61 
EXERCISES 

1. Give examples of rectangles ; of squares. 

2. Construct a rectangle having two adjacent sides equal to 
5 cm. and 8 cm. respectively (use compasses ' and straight- 
edge only). , a , 

3. Construct a rectangle hav- , . 
ing the two adjacent sides equal 

to the line segments a and b in FlG - 62 

K- 62.. 



- 

4. Construct a square whose 

side is 7 cm. long. FlG> 63 

5. Construct a square a side of which is a units long (use 
line a in Fig. 63). 

SUMMARY 

94. This chapter has taught the meaning of the follow- 
ing words and phrases : angle, vertex, vertices, initial side 
of an angle, terminal side of an angle, right angle, straight 
angle, perigon, acute angle, obtuse angle, reflex angle, 
circle, center, circumference, radius, diameter, radii, arc, 
intercept, central angle, quadrant, semicircle, latitude, lon- 
gitude, degree of latitude, degree of longitude, minute, 
second, size of an angle, protractor, degree of, arc, degree 
of angle, adjacent angles, exterior sides of an angle, field 



7^ (JKNERAL MATHEMATICS 

protractor, transit, perpendicular bisector, perpendicular to 
a line, altitude of a triangle, median of a triangle, bisector 
of an angle, parallel lines, corresponding angles, transversal, 
parallelogram, rectangle, and square. 

95. The following symbols have been introduced: Z for 
angle ; rt. /. for right angle ; A for angles ; v for arc ; 
_L for is perpendicular to ; II for is parallel to ; for dei/n-f 
or degrees ; ' for minute or minutes ; " for second or seconds. 

96. The following notations have been discussed : (1) no- 
tation for denoting and reading angles; (2) notation for 
denoting a circle by its center. 

97. This chapter has presented the important methods of 

1. Classifying angles. 
' 2. Measuring angles. 

3. Comparing angles. 

4. Drawing angles containing any amount of turning or 
any number of degrees. 

5. Adding and subtracting angles. 

6. Measuring angles out of doors. 

98. In this chapter the pupil has been taught the follow- 
ing fundamental constructions : 

1. To draw a circle. 

2. To draw an angle equal to a given angle. 

3. To draw a line perpendicular to a given line at a 
given point. 

4. To draw the perpendicular bisectors of the sides of 
a triangle. 

5. To draw the medians of a triangle. 

6. To draw a line perpendicular to a given line from a 
given point outside the line. 

7. To draw the altitudes of a triangle. 



PBOPEKTIES OF ANGLES 73 

8. To bisect a given angle. 

9. To draw the bisectors of the angles of a triangle. 

10. To draw a line through a given point parallel to a 
given line. 

11. To construct a parallelogram. 

12. To construct a rectangle. 
18. To construct a square. 

99. This chapter has taught the pupil to use the follow- 
ing instruments and devices : tracing paper, the protractor, 
and the field protractor. 

IMPORTANT GEOMETRIC RELATIONS 

100. Radii of the same circle or of equal circles are equal. 

101. In the same circle or in equal circles equal central 
angles intercept equal arcs on the circle. 

102. In the same circle or in equal circles equal arcs on 
the circle are intercepted by equal central angles. 

103. A central angle is measured by the arc intercepted 
by its sides. 

IMPORTANT DEFINITIONS 

. 104. An angle is the amount of turning made by a line 
rotating about a fixed point in a plane. 

105. A circle is a closed curve all points of which lie in 
the same plane and are equidistant from a fixed point. 

106. (1) A quadrilateral whose opposite sides are parallel 
is a parallelogram. (2) A rectangle is a parallelogram in 
which one interior angle is a right angle. (3) A square is a 
rectangle with all sides equal. 



CHAPTER IV 




FIG. 64. 



THE EQUATION APPLIED TO AREA 

107. Measuring areas. If we determine the amount of 
area inclosed within a polygon, as in the triangle ABC 
in Fig. 64, we are measuring the area of the triangle. As in 
measuring length, the process is 

one of comparison. We compare 

the area of the given polygon 

with some standardized (defined 

and accepted) unit of area and 

determine how many units are contained in the polygon ; 

that is, we determine the ratio between the area of the 

given polygon and a standard unit of area. 

108. Unit of area. The unit of area is a square each 
of whose sides is a standard unit of length. Such a unit 
involves length and width. Thus, we may 

measure area and express the result in 
square feet, square inches, square meters, 
square centimeters, etc. 

109. Practical method of estimating area. 
A practical way to estimate the area of a 
polygon is to transfer it to squared paper 

by means of tracing paper and then count the num- 
ber of square units inclosed within the figure. If the 
bounding lines cut the squares, it becomes necessary to 
approximate. In such approximations we should be careful, 
but we should not go beyond reasonable limits of accuracy. 

74 




1cm. 



1 cm. 



FIG. 65. UNIT OF 
AREA ix THE MET- 
RIC SYSTEM 



THE EQUATION APPLIED TO AREA 



75 



EXERCISES 

1. The six figures in Fig. 66 were transferred by means of 
tracing paper. Estimate the areas of each of them by counting 
the squares. Express the areas either as square centimeters 
or as square millimeters. 

HINT. One small square equals 4 sq. mm. 



FIG. 66. ESTIMATING AREAS BY MEANS OF SQUARED PAPER 

2. If the paper were ruled much finer, would you get a 
more accurate estimate ? Give an argument for your answer. 

3. Do you think that any of your results are accurate? 

4. Write a paragraph in 
precise terms, supporting your 
answer to Ex. 3. 



C- 



110. Area of a rectangle. 
CASE I. The sides of the rec- 
tangle that has been trans- 
ferred to the squared paper of 
Fig. 67 are integral multiples 
of 1 cm. Using 1 sq. cm. as a unit, there are two rows of 



FIG. 67. How TO FIND THE AREA 
OF A RECTANGLE 



76 



GENERAL MATHEMATICS 



units, and four units in a row. Counting, we see that the 
area equals 8, or 2 x 4. The law in this case is : The area 
equals the base times the altitude. In equation form this 
law may be written A = b x a. 

111. Area of a rectangle. CASE II. Let us suppose that 
we are given a square whose sides are not integral multiples 
of 1 cm. : for example, a rectangle whose base (length) 
is 2.3 cm. and whose altitude (width) is 1.3 cm. If we 
assume that the preceding law holds, then we ought to 

get 2.3 x 1.3 = 2.99 sq. cm. Instead 

of putting the rectangle on the kind 

of squared paper used in Case I, 

let us draw it again, by means 

of tracing paper, on squared paper 

that is ruled to a smaller unit, 

the millimeter, as in Fig. 68. Since 

there are 23 mm. in 2.3 cm. and 

13 mm. in 1.3 cm., if we temporarily adopt the square 

millimeter as a unit of area, then the sides of the rectangle 

are, as in Case I, integral multiples of the unit of length 

(in this case the millimeter). Hence there are 13 rows 

of units with 23 in a row, or 299 sq. mm. But there are 

100 sq. mm. in 1 sq. cm. ; hence, dividing 299 by 100, the 

result is 2.99 sq. cm., which is precisely the same number 

as that obtained by assuming the law of Case I. 

112. Area of a rectangle. CASE III. This process of 
temporarily adopting a smaller square can be continued. 
If, for example, the base of a rectangle is 2.13 cm. and 
the altitude 1.46 cm., we may imagine the rectangle to be 
drawn upon squared paper still finer ruled, that is, ruled 
to 0.1 of a millimeter. From here the reasoning is the 
same as in Cases I and II. 



FIG. 



THE EQUATION APPLIED TO AREA 



77 



EXERCISES 

1. Finish the reasoning of the foregoing paragraph. 

2. The base of a rectangle is 3^ cm. and its altitude is 5-| cm. 
Show that the area may be found by counting squares. 

3. The base of a rectangle is 5| cm. and its altitude is-2| cm. 
What unit would you temporarily adopt to find the area? 
Express the area in square centimeters. 

The preceding exercises show that if the sides of a 
rectangle involve fractions which may be expressed as 
exact decimal parts of a unit, the problem is the same 
as in Case II. 

113. Second method for finding area of a rectangle. It is 

possible to show that the transfer of a rectangle to the 

squared paper by means of tracing paper was unnecessary. 

Suppose we are given a rec- 

tangle AS CD (see Fig. 69) 

whose base is 4 cm. and 

whose altitude is 3 cm. We 

wish to find the area. Draw 

a perpendicular line to the 

line AB at the end of each 

unit segment ; that is, at the 

points E, F, and G (review 

the method of Art. 80). 

Also construct perpendiculars to the line AD at the points 

H and L Then each small square is a unit of measure 

(by definition), and the figure is divided into three rows 

of units with four in a row. By counting, the area equals 

12 (that is, base times altitude'). 



II 
I 

A 


























E F G 
FIG. 69 



78 GENERAL MATHEMATICS 

EXERCISES 

1. The base of a rectangle is 4.3cm. long, and its altitude 
is 1.7 cm. Show how to find the area of this rectangle by count- 
ing, but without the use of squared paper. 

2. Apply the law A = b x . What advantage has this law 
over the method of Ex. 1 ? 

114. Formula. An equation which expresses some prac- 
tical rule from arithmetic, the shop, the trades, the sciences, 
the business world, etc. is called a, formula. Thus, Aaxb 
is a practical formula for finding the area of a rectangle. 
The plural of "formula" is "formulas" or "formulae." 

115. Formula for the area of a square. The square is 
a special case of a rectangle ; that is, it is a rectangle in 
which a = b. The formula can be developed by the same 
method as for a rectangle. The only difference in the 
reasoning is that in every case there are as many rows of 
square units as there are square units in a row. (Why ?) 
Hence the formula for the area of a square is A = b x b. 
This formula may be written A = J 2 , where i 2 means b x b, 
and the formula is read "A equals b square." 

EXERCISES 

1. By the method of counting squares find the area of a 
square whose side is 2|- cm. long. 

2. Apply the formula A = Z/ 2 to the square in Ex. 1. .Com- 
pare results. 

3. Find the area of a square whose side is oft.; a feet; 
x inches ; y yards ; m meters ; 0.07 mm. ; 2.41 m. 

4. How many feet of wire fencing are needed to inclose a 
square lot whose area is 4900 sq. ft. ? b 2 sq. ft. ? 4 r 2 sq. yd. '.' 



THE EQUATION APPLIED TO AREA 79 

5. Express by an equation the area A of a rectangle that 
is 8 in. long and 5 in. wide ; 8 in. long and 4 in. wide ; 8 in. 
long and 3 in. wide ; 8 in. long and 6^ in. wide. 

6. Express by an equation the area A of a rectangle 12 in. 
long and of the following widths: 6 in.; 8^ in.; 9^ in.; 10| in.; 
x inches ; y inches. 

7. A mantel is 54 in. high and 48 in. wide. The grate is 
32 in. high and 28 in. wide. Find the area of the mantel and 
the number of square tiles contained in it if each tile is 3 in. 
on a side. 

8. How many tiles 8 in. square are needed to make a walk 
60 ft. long and 4 ft. wide ? 

9. Express by equations the areas of rectangles 1 in. long 
and of the following widths : 12 in. ; 9 in. ; h inches ; n inches ; 
x inches ; a inches. 

10. Express by equations the areas of rectangles of width w 
and of the following lengths : 8 ; 10 ; 12^ ; x ; a ; I ; b ; z. 

11. In each case write an equation for the other dimension 
of the rectangle, having given (a) altitude 8 in. and area 
32 sq. in. ; (b) altitude 5 ft. and area 7^ sq. ft. ; (c) base 9 ft. 
and area 30 sq. ft. ; (d) base 6 in. and area 27 sq. in. ; (e) base 
3 in. and area A square inches ; (f) base 5 in. and area A square 
inches ; (g) altitude a inches and area A square inches ; (h) base 
b inches and area A square inches. 

116. Formula for the area of a parallelogram. Fig. 70 
shows a parallelogram that has been transferred to that 
position by means of tracing paper. We wish to find its 
area. The line AB is produced (extended), and perpendic- 
ulars are dropped from D and C to the line AB (see Art. 84 
for method of constructions), thus forming the triangles 
AED inside and BFC adjoining the given parallelogram. 



80 



GEXKi; A L MATHEMATICS 



EXERCISES 

(Exs. 1-7 refer to Fig. 70) 

1. Estimate by count the number of square units in the 
triangle A ED. 

2. Estimate the number of square units in the triangle BFC. 

3. Compare the results of Exs. 1 and 2. 

4. If the area of the triangle .BFC equals the area of the 
triangle A ED, what is the relation between the area of the 
rectangle CDEF and the area of the parallelogram AB CD? 



- 



FlG. 70. HOW TO FIND THE AREA OF A PARALLELOGRAM BY 
MEANS OF SQUARED PAPER 

5. What is the formula for the area of the rectangle CDEF? 
Write the formula. 

6. What seems to be the relation between the base of the 
parallelogram and the base of the rectangle ? What evidence 
have you to support your answer ? 

7. What is the relation between the altitude of the paral- 
lelogram and the altitude of the rectangle ? Give the evidence. 

8. What seems to be the formula expressing the area of 
a parallelogram ? 

9. Without using squared paper construct a parallelogram 
(use ruler and compasses and follow the method of Art. 91). 
Divide the parallelogram into two parts a triangle and a 



THE EQUATION APPLIED TO AREA 81 

quadrilateral (as in Fig. 71). Now shift the triangle to the 
other side so as to form a rectangle. Show that the rectangle 
is equal to the parallelogram. 

The preceding exercises furnish evi- 
dence to support the following law : 
The area of a parallelogram equals the 
product of its base and altitude. This law may be written 
in the form of the following well-known formula: 

A = a x b. 

EXERCISE 

Find the area of a parallelogram if b ' = 17 in. and a = 5.3 in. ; 
if b = 15.4 in. and a 9.2 in. 

117. Area of a rhombus. The rhombus (Fig. 72) is a 
special case of the parallelogram, as it is a parallelogram 
with all its sides equal. Hence its area 

equals its base times its altitude. 

118. The area of a triangle. The ex- 
ercises that follow will help the pupil 

to understand the formula for the area of a triangle. 

EXERCISES 

(Exs. 1-12 refer to Fig. 73) 

1. Draw a triangle ABC as shown in Fig. 73. 

2. Through C draw a line CD parallel to AR (review the 
method of Art. 89). 

3. Through B draw a line parallel to AC, meeting CD at D. 

4. What kind of a quadrilateral is the figure ABDC? Why ? 

5. By means of tracing paper transfer the parallelogram 
to squared paper. 





82 GENERAL MATHEMATICS 

6. Estimate the number of square units in triangle ABC. 

7. Estimate the number of square units in triangle CBD. 

8. Compare the results of Exs. 6 and 7. What relation 
does the triangle bear to t^e parallelogram ? 

9. What seems to be the relation between the base of the 
triangle and the base of the parallelogram ? Why ? 

10. What is the relation between the altitude of the tri- 
angle and the altitude of the parallelogram ? Explain' why. 

11. What is the formula 
then for the area of any 
parallelogram ? 

12. What appears to be 
the formula for the area 

of a triangle? 

FIG. 73. How TO FIND THE AREA 

13. Construct a paral- OF A TRIANGLE 

lelogram ABCD. Construct 

the diagonal AC (a line joining opposite vertices). With a 
sharp knife cut out the parallelogram and cut along the 
diagonal so as to form two triangles. Try to make one triangle 
coincide with the other. 

14. What conclusion does the evidence of Ex. 13 support ? 

The preceding exercises furnish evidence to show that 
the area of a triangle is equal to one half the product of 
its base and altitude. This law may be written in the 
form of the following formula: 

ab 



I 
119. Area of a trapezoid. A quadrilateral having only 

two sides parallel is called a trapezoid (Fig. 74). The 
parallel sides are said to be its bases. In Fig. 74 the upper 




THE EQUATION APPLIED TO AREA 83 

base of the trapezoid is 6, the lower base is a, and the 
altitude is h. To find the area draw the diagonal BD. 

The area of the triangle ABD = -.h. Why ? 

The area of the triangle BCD = --h. Why ? 

2,7 

Therefore the area of the trapezoid = a '- + b - Why ? 

77 2t 2 

Note that a and b - are similar terms. Why ? In the 

h 
first term a is the coefficient of and in the second term 

b is the coefficient; hence, adding coefficients, as we may 
always do in adding 
similar terms, the area 

of trapezoid is (a + b) - 

We can only indicate A 

the sum of the two FlG - 74 ' How J FIND THE AREA OF 

A TRAPEZOID 

bases until we meet an 

actual problem. The parenthesis means that a + b is to 
be thought of as one number. The law is: The area of 
a trapezoid is equal to one half the product of its altitude by 
the sum of its bases. This law may be written in the form 
of the following formula: 



EXERCISES 

1. Find the area of the trapezoid whose altitude is 12.6 in. 
and whose bases are 8 in. and 4.6 in. respectively. 

2. The altitude of a parallelogram is 3 x + 2, and its base 
is 4 in. Write an algebraic expression representing its area. 
Find the value of x when the area is 28 sq. in. 



84 



GENERAL MATHEMATICS 



D 



70 




FIG. 75 



3. The altitude of a triangle is 10 in., and the hast- is 
3 x + 2 in. Write an algebraic number representing the area. 
Find the value of ./ when the area is 55 sq. in. 

4. A man owns a city lot with the form 
and dimensions shown in Fig. 75. ^He wishes 
to sell his neighbor a strip AEFD having a 
frontage DF equal to 10 ft. If the property 
is worth $5600, how much should he receive 
for the strip '.' 

5. Of what kinds of polygons may the 
following equations express the areas '.' 

(a) A=x\ (h) A=l 

(b) A = 3.*-. _3 

(d) -i =5 (a: + 3). 4 

(e) .4 = (. + 2). 

(f) .4 = a(5 + 4), 

(g) A = *(y + 2). 

6. Find the value of ,1 in Ex. 5 when x = 3, y = 2, a = 4, 
b = 1, and c = 5. 

7. What quadrilaterals contain right angles ? 

8. In what respect does the square differ from the rectangle ? 

9. Having given a side, construct a square, using only ruler 
and compasses. 

HINT. Review the method for constructing a perpendicular to a 
line segment (Art. 80). 

10. Hg>w does a square differ from a rhombus ? 

11. Is a rhombus a parallelogram ? Is a parallelogram a 
rhombus ? 

12. Construct a rhombus with ruler and compasses, given a 
side equal to 5 cm. and given the included angle between two 
adjacent sides as 41. 

HINT. Use the construction for parallel lines (Art. 89). 



THE EQUATION APPLIED TO AREA 



85 



GEOMETRIC INTERPRETATION OF PRODUCTS 

120. A monomial product. The formulas for the area of 
the rectangle, the square, the triangle, the trapezoid, etc. 
show that the product of numbers 
may be represented geometrically ; 
for example, the product of any 
two numbers may be represented 
by a rectangle whose dimensions FIG. ~ 6 - ILLUSTRATING A 
are equal to the given numbers. . 
Thus the rectangle in Fig. 76 represents the product ab. 



ab 



EXERCISES 

1 . Sketch a rectangle to represent the product 6 x. 

2. Sketch an area to represent kry. 

3. Show from Fig. 77 that the area 
4 x 2 . What is the product of 2 a- x 2x? 

4. Show by means of a figure the area 
of a rectangle 3 a by 5 a. 

5. Draw a figure to represent the prod- x 
uct of 5 a- and 4 x. 

6. On squared paper draw an area repre- 

lio. 77. ILT.USTRAT- 
senting the product ab. To the same scale 1NG THE SQL-ARE OF 

draw the area ba. Compare the areas. A MONOMIAL 

7. Show by a drawing on squared paper that 4-5 = 5-4. 

121. Law of order. The last two exercises illustrate 
that in algebra, as in arithmetic, the fen 'torn of a product 
may be changed in order without changing the value of the 
product. Thus, just as 2x3x5 = 5x3x2, so xyz zyx. 
This is called the Commutative Law of Multiplication. 



i. 

x 

x 


3 expressed 
x x 


by 

X 

X 






x* 




X X 



GENERAL MATHEMATICS 



EXERCISE 

Simplify the following : (a) 2 x 3 y . 4 z ; (b) (2 x y) (3 x y} ; 
(c) 4 x 2/ 3 x my. 

122. Product of a polynomial and a monomial. The 

formula for the trapezoid suggests the possibility of draw- 
ing areas to represent the product of a sum binomial by 
a monomial. The process is illustrated by the following 
exercises. 

EXERCISES 

1. Express by means of an equation the area of a rectangle 
of dimensions 5 and x + 3 (see Fig. 78). The area of the 
whole rectangle equals 5 (x -\- 3). D A F 

Why? If a perpendicular be erected 
at B (see Art. 80 for method), the 5 
rectangle is divided into two rec- 
tangles. The area of DCBA equals 
5x. Why? The area of ABEF 

equals 15. Why? It is now easy PRODUCT OF A POLYNOMIAL 

f. T ,, , AND A MONOMIAL 

to hnd the entire area DCEF. 



5x 



15 



C X B 3 E 

FIG. 78. ILLUSTRATING THE 



Since DCEF = DCBA + ABE1*, 

T> (x + 3) =5 x + 15. 

2. Show from Fig. 79 that a(x + ?/) = ax + "//. 

3. Show by Fig. 80 that a (x + T/ + z) = ax + a?/ + az. 



Why ? 
Why? 







x y 
FIG. 79 









X y Z 
FiG. 80 



4. Draw an area to represent bm + In + be. 

5. Draw an area to represent 2 ax + 2 ay + 2 az. 

6. Kepresent 2cc + 4?/ + 6byan area. 

7. Sketch a rectangle whose area equals 2 ax + 2 ay + 6 az. 




EUCLID 



88 GENERAL MATHEMATICS 



HISTORICAL NOTE. The word "geometry" comes from a Greek 
phrase which means to measure the earth. The early Egyptians had 
serious need for a reliable method of measuring the land after each 
overflow of the Nile. The early history of geometry appears to rest 
on this practical basis. 

The oldest collection of geometry problems is a hieratic papyrus 
written by an Egyptian priest named Ahmes at a date considerably 
earlier than 1000 H.C., and this is believed to be itself a copy of some , 
other collection a thousand years older. 

Ahmes commences that part of his papyrus which deals with 
geometry by giving .some numerical instances of the contents of 
barns. Since we do not know the shape of the barns, we cannot 
check the accuracy of his work. However, he gave problems on 
pyramids. The data and results given agree closely with the dimen- 
sions of the existing pyramids. 

Geometry took definite form as a science when Euclid (about 
300 B.C.) wrote his "Elements of Geometry." The proofs of his 
text were so excellent that the book replaced all other texts of 
the time and has held an influential position to this day. The 
form of Euclid is practically the same as most American geom- 
etry texts, and in England boys still say they are studying Euclid 
(meaning geometry). 

We know little of Euclid's early life. He may have studied in 
the schools founded by the great philosophers Plato and Aristotle at 
Athens, in Greece. He became head of the mathematics school 
at Alexandria, Egypt, and proceeded to collect and organize into a 
set form the known geometric principles. He is said to have insisted 
on the knowledge of geometry for its own sake. Thus, we read of 
his telling the youthful Prince Ptolemy, " There is no royal road to 
geometry." At another time, so the story goes, when a lad who had 
just begun geometry asked, "What do I gain by learning all this 
stuff ? " Euclid made his slave give the boy some coppers, " since," 
said he, " he must make a profit out of what he learns." 

Euclid organized his text so as to form a chain of reasoning, begin- 
ning with obvious assumptions and proceeding step by step to results 
of considerable difficulty. The student should read about his work in 
Ball's "A Short Historyof Mathematics." Cajori's" History of Elemen- 
tary Mathematics " and Miller's " Historical Introduction to Mathe- 
matical Literature " are further sources of information about Euclid. 



THE EQUATION APPLIED TO AREA 89 

123. Partial products. The products ax, ay, and az in 
the polynomial ax + ay 4- az are called partial products. 
Each term of such an expression may be used to represent 
the area of some part of a rectangle. For example, Fig. 80 
shows a rectangle divided into the parts ax, ay, and az 
respectively. Here the polynomial ax + ay + az may be 
said to represent the area of the whole rectangle. 

124. Algebraic multiplication. The list of exercises in 
Art. 122 also shows that the product of a polynomial and a 
monomial is found by multiplying each term of the polynomial 
by the monomial and then adding the partial products. 

EXERCISES 

1. Perform the following indicated multiplications : 

(a) 3(2 x + 3 ij). (c) 3c(2 + 36). 

(b) (5x + 2z)4:a. (d)3e(2c + 3). 

2. Letting x = 3, y = 1, and z = 2, find the value of the 
following numbers : (x + y)z; 2 (x -j- ?/) 2 ; 3 x + 2 (y + z) ; 

2z); 2s + 5(x + ,/). 



125. Geometric product of two polynomials. The product 
of a + c and x + y may be indicated as (a + c) (x + y}. 
This product may be represented 






cy 






, . ,, , <-,.,.. , ~ ay 

geometrically (rig. ol) by a rec 

tangle whose base is x + y and 
whose altitude is a + c. FIG. 81. ILLUSTRATING 

The rectangle is composed of. THE PRODUCT OF TWO 
four rectangles: ax, ay, ex, and cy. 

By the axiom that the whole is equal to the sum of its parts, 
the whole rectangle, ( + <?)(#+?/), equals ax+ay + cx + cy, 
the sum of the parts. 



90 GENERAL MATHEMATICS 

EXERCISES 

1: Sketch a rectangle whose area will be the product of 
(a + ft) (e + d). 

2. Find the geometric product of (c + t) (m + n). 

3. Perform the multiplication (2 -f x)(m -f w) by means of 
a geometric figure. 

4. Find the product (3 x + 2 ?/) (a + b). Sketch the area 
represented by this product. 

5. Find the product (a + b~)(x + ;/ + z), using a geometric 
figure. 

126. Algebraic product of two polynomials. The figures 
drawn in the preceding exercises indicate a short cut in 
the multiplication of two polynomials. Thus, a polynomial 
is multiplied by a polynomial by multiplying each term of 
one polynomial by every term of the other and adding the 
partial products. 

EXERCISES 

1. Using the principle of Art. 126, express the following 
indicated products as polynomials : 

(a) (m + n) (a + 5). (g) 3(2 a 2 + a + 5). 



(f) 5(4 +7 + 3). (j) (3x 

(k) (5 b + 2 c + 3 d) (2 x + 3 y + 4 z). 
(1) (2 m + 3 n -f- p] (3 a + 7 J + 5*). 



2. One side of a rectangle is 4yd. and the other is 6yd. 
How much wider must it be made so as to be l times as 
large as before? / 



THE EQUATION APPLIED TO AREA 91 

3. Multiply the following as indicated and check the result : 



Solution. 2 x (x + 2 y + y-) = 2 x 2 + 4 xy + 
Check. Let x = 2 and y 3. 



2 x 2 + 4 xy + 2 xy" = 8 + 24 + 36 = 68. 

NOTE. Avoid letting x = 1, for in this case 2 x, 2 x 2 , 2 x 8 , etc. are 
each equal to 2. Why? 

4 . Multiply the following as indicated and check the results : 
(a) 

(b) 

(c) (m + n + a) (m + n + &). 

(d) (m + n + 2 

(e) (0.4 x + 0.3 y + 0.6 s) (10 x + 20 y + 30 z). 

127. Geometric square of a binomial. The product of 
(x + y*)(x + yy, or (z + y) 2 , is an interesting special case 
of the preceding laws. The prod- 
uct may be represented by a square 
each of whose sides is x + y (see 
Fig. 82). The square is composed of 
four parts, of which two parts are 
equal. Since these two parts are repre- 
sented by similar algebraic terms, they 
may be added ; thus, xy -\-xy-2 xy. x V 

Hence the area of a square whose F IG. 82. ILLUSTRATING 
j o . r> o rru THE SQUARE OF A Bi- 

side isaj-t-yisar+Jsa^ + ^r. 1 he 

same product is obtained by apply- 
ing the law for the product of two polynomials ; thus, 

x +y 

x +y 



xy 



xy 



+ 



xy 

xy + 



2 xy 



92 GENERAL MATHEMATICS 

In algebraic terms we may say that the square of the 
sum of tivo numbers equals the square of the first, plus twice 
the product of the two numbers, plus the square of the second. 

Use Fig. 82 to show what this law means. 

EXERCISES 

1. By means of figures express the following squares as 
polynomials : 



(b) (m + n) 2 . (e) (x + 2) 2 . (h) (2 x + y) 2 . 

(c) (c + d)*. (f ) (m + 3f. (i) (2 x + 3 y) 2 . 

2. Sketch squares that are suggested by the following 
trinomials : 

(a) a 2 + 2 ah + lr. (e) m z + 8 MI + 16. 

(b) x 2 + 2 ax + a 2 . (f) .r 2 + 10 a- + 25. 

(c) A- 2 + 2 A->- -(- r. (g) 49 + 14 x + z 2 . 

(d) x 2 + 4* + 4. (h) c 2 + c + ], 

3. Indicate what number lias been multiplied by itself to 
produce 

(a) or 2 + 2 vy + /. (c) z 2 + 6 a- + 9. 

(b) r 2 + 4 r + 4. (d) & a + 10 // + 25. 

4. What are the factors in the trinomials of Ex. .3 '.' 

5. The 'following list of equations review the fundamental 
axioms as taught in Chapter I. Solve each equation and check 
by the methods of Chapter I. 

(a) 30 + 4) =22 + ,-. 

(b) 9 + 35) =5 (2 a + 45). 

(c) 3(x + 15)+ 5 = 2(2* + 9) + 4(.r + 3). 

(d) ?fe2-8. W |i-| = 1 . (h)f-f=a 

2 . fc, 5 + 5-2. (i)^^ = 8. 



. 



93 

128. Evaluation. The area of each of the geometrical 
figures considered in this chapter has been found to depend 
upon the dimensions of the figure. This dependence has 
been expressed by means of formulas, as A = ab in the case 
of the rectangle. Whenever definite numbers are substi- 
tuted in the expression ab in order to find the area, A, 
for a particular rectangle, the expression ab is said to be 
evaluated. This process implies getting practical control 
of the formulas. 

EXERCISES 

1. Find the value of A in the formula .1 = ab when 
a = 22.41 ft. and b = 23.42 ft. 

2. Find the value of J in the formula A = when 
a = 12.41 ft. and I = 2.144 ft. 

3. Find the value of .1 in the formula A = (<i -)- fi) when 
a = 12.42 ft, b = 6.43 ft., and h = 20.12 ft. 

129. The accuracy of the result. In finding the area in 
Ex. 1 above we get A = (22.41) (23.42)= 524.8422 sq. ft. 
This is a number with four decimal places. As it stands 
it claims accuracy to the ten-thousandth of a square foot. 
The question arises whether this result tells the truth. 

Suppose the numbers above represent the length and width respec- 
tively of your classroom. Does the product 524.8422 sq. ft. indicate 
that we actually know the area of the floor accurate to one ten- 
thousandth of a square foot? Shall we discard some of the deci- 
mal places ? If so, how many are meaningless ? How much of the 
multiplication was a waste of time and energy ? These questions are 
all involved in the fundamental question How many decimal places 
shall we regard as significant in the process of multiplication ? . 

It is important that we have a clear understanding of 
the question. For if we carry along in the process mean- 
ingless decimals we are wasting time and energy, and, 



94 

what is more serious, we are dishonestly claiming for 
the result an accuracy which it does not have. On the 
other hand, we are not doing scientific work when we 
carelessly reject figures that convey information. 

The following facts are among those which bear on our 
problem : 

(a) In Art. 26 we pointed out that any number obtained 
by measurement is an approximation.' The application of 
the area formulas involves the measurement of line seg- 
ments. Hence an area is an approximation. This fact 
alone is sufficient to make us exceedingly critical of the 
result 524.8422 sq. ft. as an absolutely accurate measure 
of the area of the classroom floor. 

(b) If we measure the length of a room with a reliable 
tape measure and record the result as 23.42 ft., this does 
not mean that we regard the result as absolutely accurate. 
If the scale is graduated to hundredths of a foot, it means 
that 23.42 ft. is the result nearest to the true value. The 
eye tells us that 23.425 ft. is too high and 23.415 ft. is 
too low, but that the result may be anywhere between 
these. Thus, the length of the room lies anywhere between 
23.415 ft. and 23.425 ft. Similarly, the width may be 
anywhere between 22.405 ft. and 22.415 ft. The student 
should practice measuring objects with a yardstick or a 
meter stick till the point of this paragraph is clear to him. 
Test question : How does 2.4 ft. differ from 2.40 ft. ? 

Multiplying the smallest possible length (23.415 ft.) 
of the classroom by the smallest width (22.405 ft.) we 
get a possible area of 524.613075 sq. ft. By multiplying 
the greatest length (23.425 ft.) by the greatest width 
(22.415 ft.) we get 525.071375 sq. ft. Subtracting the 
smallest possible area from the largest possible area gives 
us a range of over 0.45 of a square foot. In short, the 



THE EQUATION APPLIED TO AREA 95 



result might be wrong by practically one half of a square 
foot. We are not actually sure of the third figure from the 
left. It mayfce a 4 or a 5. We shall be reasonably near 
the truth if w% record the result simply as 524.? sq. ft., 
a number chosen roughly halfway between the largest and 
smallest possible areas. 

It can thus be shown that the product of two approxi- 
mate four-place numbers is not to be regarded accurate 
to more than four places. 

*130. Abbreviated multiplication. It is apparent in the 
preceding discussion that it is a waste of time to work 
out all the partial products in multiplication. It is easier 
(when the habit is once established) to work out only 
the partial products which go to make up the significant 
part of the answer. 

Thus, 47.56 x 34.23 may take the following forms : 

ABBREVIATED FORM USUAL METHOD 

47.56 By multiplication we get 

34.23 47.56 

1427 34.23 

190 
10 
1 
1628. 



The difference is accidentally only a little more than 
0.02 sq. ft. It can be shown by the method used in the 
classroom problem (Art. 129) that 1628 is easily in the 
range of probable areas ; that is, we are not actually sure 
about the fourth figure from the left. 

* Hereafter all articles and exercises marked with an asterisk may be 
omitted without destroying the sequence of the work. 



1 


4268 


9 


512 


190 


24 


1426 


8 


1627 


9788 



96 GENERAL MATHEMATICS 

The abbreviated method consists of writing only the 
significant parts of the usual method (see numbers to left 
of the line). Add 1 unit when the number^o the right is 
the figure 5 or larger. The method will pppear awkward 
until sufficiently practiced. 

A similar discussion concerning accuracy could be given 
for division. In addition or subtraction it is easy to 
see that the sum or difference of two numbers cannot be 
regarded as more accurate than the less accurate of the 
two numbers. Illustrate the truth of the last statement. 

While the discussion of this very important topic has 
been by no means complete, perhaps enough has been 
said to fulfill our purpose, which is to make the student 
exceedingly critical of results involving the significance of 
decimal places. 

EXERCISES 

*1. Assuming that the dimensions of a hall are measured 
with a reliable steel tape and that the dimensions are recorded 
as 47.56 ft. and 34.23 ft. respectively, show by the method used 
in Art. 129 that the difference between the smallest and the 
largest possible area of the hall is actually over four fifths of 
a square foot. 

*2. By means of the abbreviated multiplication method 
write the product of 46.54 and 32.78 ; of 23.465 and 34.273. 

*3. Multiply by the usual method and compare the short- 
cut result with this result. 

*4. Which result is the more accurate ? 

SUMMARY 

131. This chapter has taught the meaning of the follow- 
ing words and phrases : area, measuring area, unit of area, 
rhombus, trapezoid. Commutative Law of Multiplication, 
partial products, parenthesis, formula, formulas. 



THE EQUATION APPLIED TO AREA 97 

132. The following formulas have been taught: 

(a) A = ba. (For the area of a rectangle.) 

(b) A = b 2 . (For the area of a square.) 

(c) A=bh. (For the area of a parallelogram.) 

(d) A = (For the area of a triangle.) 

7 

(e) A = (a + >) - (For the area of a trapezoid.) 

a 

(f ) (x + #) 2 = y?+ 2 xy + 3/ 2 . (For the area of a square 
whose side is x 



133. The product of two numbers may be represented 
geometrically as an area. 

134. The algebraic product of a monomial and a poly- 
nomial is found by multiplying each term of the polynomial 
by the monomial and then adding the partial products. 

135. The product of two polynomials is the sum of all 
the partial products obtained by multiplying each term 
of one polynomial by every term of the other. 

*136. We need to be very critical of the number of 
decimal places that we submit in a result. The product of 
two approximate four-digit numbers is only approximately 
correct for four digits. 



CHAPTER V 

THE EQUATION APPLIED TO VOLUME 

137. Solids. The drawings in Fig. 83 represent geo- 
metric solids. A solid is commonly thought of as an object 
that occupies a portion of space. It is separated from 





Oblique Paral- 
lelepiped 



Cube 



Rectangular Paral- 
lelepiped 






Triangular 
Pyramid 



Sphere 
FIG. 83. FAMILIAR SOLIDS 



Frustum of a 
Pyramid 



the surrounding space by its surface. In geometry we 
study only the form of the solid and its size. We are 
not interested in color, weight, etc. A solid differs from 
the figures we have been studying in that it does not lie 
altogether in a plane, but involves a third dimension. What 
figures in two dimensions are suggested by the solids in 
Fig. 83 ? For example, the square is suggested by the cube. 

98 



THE EQUATION APPLIED TO VOLUME 



99 



138. Cube. The cube has six faces all of which are 
squares. Two faces intersect in an edge. How many edges 
has a cube ? How many corners ? How is a corner formed ? 

139. Oblique parallelepiped. The faces of an oblique 
parallelepiped are all parallelograms. How many faces 
has it? How many vertices? How many edges? 

140. Rectangular parallelepiped. The faces of a rec- 
tangular parallelepiped are rectangles. 

141. Measurement of volume ; unit of volume. When we 
determine the amount of space inclosed within the surface 
of a solid we are measuring the volume of the solid. To 
measure the volume of a solid we compare the solid with 
a cube each of whose edges equals a unit of length. The 
volume is expressed numerically by the number of times 
the unit cube goes into the solid. The unit cube is called 
the unit of volume. 

142. Formula for the volume of a rectangular parallele- 
piped. In Fig. 84 a rectangular parallelepiped is shown 
which is 5 cm. long, 3 cm. wide, 

and 4 cm. high. The unit cube 

is represented by K. Since the 

base of the solid (the face 

on which it stands) is 5 cm. 

long and 3 cm. wide, a layer of 

3 x 5 unit cubes could be placed 

upon it. Since the solid is 4 cm. 

high, it contains 4 layers of unit 

cubes ; that is, 4x3x5, or 60, 

unit cubes. Thus the volume 

of a rectangular parallelepiped 

is obtained by multiplying the length by the width by the 

height. This law may be expressed by the formula V= Iwh. 



B 



/ / 




7 



N 



R 

FIG. 84. How TO FIND THE 
VOLUME OF A RECTANGULAR 
PARALLELEPIPED 



100 



EXERCISES 

1. Find the volume of a rectangular parallelepiped if its 
dimensions are I = 63 in., h = 42 in., and w 56 in. 

*2. If in the preceding discussion the edges of the rec- 
tangular parallelepiped had not been given as integral multiples 
of the unit cube, it would have been necessary temporarily to 
adopt a smaller unit cube. Show that the formula V = Iwh 
holds when I = 2.3 cm., h 3.4 cm., and w = 1.7 cm. 
HINT. Follow the method suggested in Art. 111. 

*3. Show by means of a general discussion that the formula 
would be true if I = 3j, h = 2j, w = 3f . 
HINT. See Ex. 3, Art. 112. 

* 143. Volume of an oblique parallelepiped. Fig. 85 shows 
in a general way the method used in a more advanced 
mathematics course to show that the formula V= Iwh holds 




in 



}h 



R 



FIG. 85. MODEL ILLUSTRATING HOW TO FIND THE VOLUME OF AN 
OBLIQUE PARALLELEPIPED 

even for aji oblique parallelepiped. Parallelepiped III is a 
rectangular parallelepiped, and we know the formula holds 
for it. Parallelepiped II is a right parallelepiped (it has 



THE EQUATION APPLIED TO VOLUME 101 

four rectangular faces, and two are parallelograms) and 
by advanced methods is shown equal to parallelepiped III. 
Parallelepiped I is oblique and is shown equal to parallele- 
piped II. Since parallelepiped I equals parallelepiped II, 
and parallelepiped II in turn equals parallelepiped III, the 
formula holds for parallelepiped I. The student should not 
be concerned if he cannot fully understand this discussion. 
He should be ready to apply the formula for an oblique 
parallelepiped when the need for it arises in shop or factory 
just as he does many principles of arithmetic. 

EXERCISES 

*1. It will be easy for some student to make models of the 
preceding figures in the shop. Thus, to show parallelepiped II 
equal to parallelepiped III construct parallelepiped II and 
drop a perpendicular from D to the base. Then saw along the 
edges MD and DF. Place the slab obtained on the right side, 
and parallelepiped II will look exactly like parallelepiped III. 
This will be helpful to your classmates, and you will find the 
exercise easy and interesting. 

*2. A much more difficult and interesting exercise is to make 
parallelepiped I look like parallelepiped III. 

HINT. Construct RK to AC and KI to AC. Saw along the 
edges RK and KI and place the slab obtained on the left side. Now 
the figure will be transformed to parallelepiped II. Continue as in 
Ex. 1 to make parallelepiped I look like parallelepiped III. 

3. A rectangular reservoir 120 ft. long and 20 ft. wide con- 
tarns water to a depth of 10.5 ft. A second reservoir 125 ft. long 
and 18 ft. wide contains water to a depth of 36.5 ft. How much 
more water is there in the. second reservoir than in the first ? 

4. A rectangular tank 6 ft. long, 4 ft. wide, and 5 ft. deep 
is to be lined with zinc ^ in. thick. How many cubic feet of 
zinc will be required if 4 sq. ft. are allowed for overlapping ? 



102 GENERAL MATHEMATICS 

5. If 1 cu. in. of pure gold beaten into gold leaf will cover 
30,000 sq. ft. of surface, what is the thickness of the gold leaf ? 

6. An open tank is made of iron ^ in. thick. The outer 
dimensions are as follows : length, 3 ft. ; width, 1 ft. 9 in. ; 
height, 2 ft. If 1 cu. ft. of iron weighs 460 lb., find the weight 
of the tank. 

7. In a rainfall of 1 in. how many tons of water fall upon 
an acre of ground if 1 cu. ft. of water weighs 62.5 lb. ? 

144. Formula for the volume of a cube. The volume of 
a cube is computed in the same way as that of a parallele- 
piped. The cube is a special case in the sense that the 
length, width, and height are all equal. Hence, if s equals an 
edge of a cube, the volume may be expressed by the formula 
V = s x s x s. The formula V= s x s x s may be written 
more briefly V= s 3 (read "V equals s cube"); s 3 being an 
abbreviated form of s x s x s. The formula may be trans- 
lated into the following law : The volume of a cube equals 
the mbe of an edge. 

EXERCISES 

1. Find the volume of a cube whose edge is ^ in. ; ^ in. ; ^ in. 

2. Find the volume of a cube whose edge is 1^ in.; 2.2 cm.; 
3 in.; 1 m. ; 0.01 m. 

145. Equal factors ; exponents ; base ; power. The prod- 
ucts of two equal dimensions and three equal dimen- 
sions have been represented by the area of a square and 
the volume of a cube respectively. Hence the notation 
" a square " and "s cube." The product of four equal fac- 
tors cannot be represented geometrically, though you may 
already have heard people talk vaguely about the fourth 
dimension. However, the product of four equal algebraic 
factors, say s x s x s x s, is as definite as 2x2x2x2 



. THE EQUATION APPLIED TO VOLUME 103 

in arithmetic. Thus, we extend the process indefinitely in 
algebra and write sxsxsxs = s* (read "s fourth") or 
bxbxbxbxb = b* (read " b fifth "), etc. 

The term b 5 is obviously much more convenient than 
bxbxbxbxb. The 5 hi b 6 is called an exponent. It 
is a small number written to the right and a little above 
another number to show how many times that number is to 
be used as a factor. In 5 3 (meaning 5x5x5) the 3 is 
the exponent, the number 5 is the base, and the product 
of 5 x 5 x 5 is the power. Thus, 125, or 5 3 , is the third 
power of 5. When no exponent is written, as in x, the 
exponent is understood to be 1. Thus, in 2 xy, both x and 
y are each to be used only once as t a factor. The mean- 
ing is the same as if the term were Mjritten 2x l y l . 

EXERCISES 

1. State clearly the difference between a coefficient and an 
exponent. Illustrate with arithmetical numbers. 

2. Letting a = 5, give the meaning and the value of each 
of the following numbers : 

(a) 3 a. (c) 2 a. (e) 4 a. (g) 5 a. (i) 2 a 2 . (k) 4 a 2 . 

(b) a 8 . (d) a?. (f) a 4 . (h) a 5 . (j) 3 a 2 . (1) 2 a 4 . 

3. Write the following products in briefest form : 

1111 333 a a a 

yyyyy, 5-5-5-5; g-^ji m'> 

2J . 5J 2J ; 5 . 4 V4 f 6 3 ^ 3 3 * . y . y . y. 

4. Find the value of 2 s ; 6 8 ; (i) 4 ; 3 8 ; (1.3) a ; 9 8 ; (0.03) 8 ; 
(1.1) 8 . 

5. Letting m = 2 and n = 3, find the value of the follow- 
ing polynomials : m 2 + 2 mn + w 2 ; m" + 3 ra 2 n, + 3 wp 2 + w 8 ; 

); 5(m + n); 6(2 w s + 3m 2 -f- 4mn + n + 3). 



104 GENERAL MATHEMATICS 

6. Find the value of the following numbers, where z 3 : 
(a) 2z; (b) *; (c) (2*) 2 ; (d) 2* 2 ; (e) 3z 8 ; (f) (3z) 3 ; 

(g) (3*) 2 - 

7. Letting a; = 1, y = 2, z = 3, and u = 4, find the value of 
the following : 

xy + xz + yu + zu x* + 4 x 3 # -f- 6 x 2 y 2 4- 4 x^ 3 + 2/ 4 
# -f y + + w xy 

146. Exponents important. Since the subject of expo- 
nents is fundamental to a clear understanding of two very 
important labor-saving devices, namely the slide rule and 
logarithms, which we shall presently study, it is necessary 
to study the laws of exponents very carefully. 

147. Product of powers having the same exponents. The 
law to be used in this type may be illustrated by the 
problem, " Multiply a 2 by a 3 ." The expression a 2 means 
a a, or aa, and the expression a? means a a a, or aaa. 
Hence a 2 a 3 means aa aaa, or, hi short, a 6 . 

EXERCISE 

In each case give orally the product in briefest form : 

(a) 3 2 -3 3 . (i) x s - x 6 . (q) x 2 m. 

(b) 6-6 8 . (j) ax-x. (r)6-6.*>fl 

(c) 5 2 5*. (k) b e b. ( s ) 4 tfc 5 iV. 

(d) 10 10 8 . (1) b-b. ( ; t ) xy- x 2 yz 2 2 xifz s . 

(e) x x 2 . (m) e e 2 2. (u) (2 xyf. 

(f) 12 2 .12 S . (n) c-c 3 . (v) (2xV) 3 . 

(g) x 2 x s . ( o ) x x 5 . (w) ' (3 x 2 ?/) 2 . 

(h) x x 4 . (p) m a- 2 . (x) 3 5 2 5 2 3 8 . 

The exercise above shows that the product of two or 
more factors having the same base is a number ivJwse base 
is the same as that of the factors and whose exponent is the 
mm of the, exponent* of the factors ; thus, b 2 & . / = 2> 10 . 



THE EQUATION APPLIED TO VOLUME 105 

148. Quotient of powers having the same base. The quo- 
tient of two factors having the same base may be simplified 
by the method used in the following problem: 

Divide b 6 by b\ 

HINT. Since a fraction indicates a division, this quotient may 

6 5 

be indicated in the form 
cr 

b 5 b-b'b-b'b 

Solution. - = - - -- 

&* b b 

Hence, dividing numerator and denominator by b b, or b 2 , 



EXERCISE 

In each case give orally the quotient in briefest form : 

2 (c)ft'-*-P. (g)* 6 ^* 2 . 

w 2" ' (d) b 6 -- b\ (h) x 6 -4- x s . 

5_ (e)b" + b*. (i)x + x*. 

W 5 2 ' (f) b + bv (j) x 21 + x". 

The preceding exercise shows that the quotient obtained 
by dividing a power by another power having the same base 
is a number whose base is the common base of the given 
powers and whose exponent is obtained by subtracting the 
exponent of the divisor from the exponent of the dividend. 

Thus, a: 10 * x 3 = x 1 . Here x 1 has the same base as the dividend x w 
and the divisor X s ; and its exponent, 7, is obtained by subtracting 3 
from 10. 

149. Review list of problems involving application of 
algebraic principles to geometric figures. The following 
exercises are intended to help the student to see how to 
apply algebraic principles to geometric figures. 



106 



2X+3 




FIG. 



EXERCISES 

1. Find an algebraic number which will express the sum 
of the edges of the solid in Fig. 86. 

2. If the sum of the edges of the solid in Fig. 86 is 172, what 
are the actual dimensions of the solid ? 

3. Find an algebraic expression 
for the total surface of the solid in 
Fig. 86. Also for the volume. 

4. What is the total surface and vol- 
ume of the solid in Fig. 86 if x equals 10 ? 

5. Express algebraically the sum of the edges of the cube 
in Fig. 87. 

6. If the sum of the edges of the culx- 
in Fig. 87 is 112, what is the length of 
one edge ? 

7. Express algebraically the total surface 
and volume of the cube in Fig. 87. 

8. What is the total surface of the cube in Fig. 87 if x = 2 ? 

9. The edge of a tetrahedron (Fig. 88) is denoted by 
2 x + 1. Express algebraically the sum of all 

the edges of the tetrahedron. 

NOTE. A tetrahedron is a figure all of whose 
edges are equal and whose faces are equal equilateral 
triangles. 

10. Find the length of an edge of the tetra- 
hedron in Fig. 88 if the sum of the edges is 40.5 cm. 

11. Fig. 89 shows a frustum of a pyra- 
mid. The upper and lower bases are equi- 
lateral pentagons ; the sides are trapezoids 
with the edges denoted as in the figure. 
Find the sum of all the edges. If e 

equals 2. what is the sum of the edges ? FIG. 89 







DESCARTES 



108 GENERAL MATHEMATICS 

HISTORICAL NOTE. The idea of using exponents to mark the 
power to which a quantity was raised was due to Rene 1 Descartes, 
the French philosopher (1596-1650). It is interesting to read of the 
struggle for centuries on the part of mathematicians to obtain 
some simple method of writing a power of a number. Thus, we read 
of the Hindu mathematician Bhaskara (1114- )*using the initials 
of the Hindu words " square " and " solid " as denoting the second and 
third power of the unknown numbers in problems, which he gave a 
practical setting with many references to fair damsels and gallant 
warriors. In the following centuries a great variety of symbols for 
powers are used ; for example, arcs, circles, etc., until we come to a 
French lawyer, Frangois Vieta (1540-1603), who wrote on mathe- 
matics as a pastime. Vieta did much to standardize the notation of 
algebra. Thus, in the matter of exponents he employed " A quad- 
ratus x " " A cubus," to represent z 2 and x 8 , instead of introducing a 
new letter for each power. From this point it is only a step to 
Descartes's method. 

The biographies of the three mathematicians Bhaskara, Viet;a, and 
Descartes are exceedingly interesting. Thus, you may enjoy reading 
of Bhaskara's syncopated algebra in verse, in which many of the 
problems are addressed to " lovely and dear Lilavati " (his daughter) 
by way of consolation when he forbade her marriage. 

You may read of Vieta's being summoned to the court of Henry IV 
of France to solve a problem which involved the 45th power of x. 
The problem had been sent as a challenge to all mathematicians 
in the empire. Vieta appeared in a few moments and gave the king 
two correct solutions. Next King Henry asked Vieta to decipher 
the Spanish military code, containing over six hundred unknown 
characters, which was periodically changed. King Henry gave the 
cipher to Vieta, who succeeded in finding the solution to the system, 
which the French held greatly to their profit during the war. 

Or you may read of Descartes, a member of the nobility, who 
found the years of his army life exceedingly irksome, for he craved 
leisure for mathematical studies. He resigned his commission in 
1621 and gave his time to travel and study. In 1637 he wrote a book, 
" Discourse on Methods." In this text he made considerable advance 
toward the system of exponents now used. The text shows that he 
realized the close relation existing between geometry and algebra. 
He is often called "the father of modern algebra." 



THE EQUATION APPLIED TO VOLUME 109 



12. A tetrahedron may be constructed from a figure like 
Fig. 90. Draw the figure on cardboard, using a larger scale. 
Cut out the figure along the heavy lines ; then fold along 
the dotted lines. Join the edges by means of gummed paper. 




FIG. 90. How TO CON- 
STRUCT A TETRAHEDRON 



FIG. 91. How TO CONSTRUCT 
A CUBE 



13. The cube may be constructed from a figure like Fig. 91. 
Draw the figure on cardboard, using a larger scale ; for example, 
let x = 3 cm. Cut out the figure 

along the heavy lines, then fold 
along the dotted lines. Join the 
edges by means of gummed paper. 
This will form a model of a cube. 

14. Measure the edge of the 
cube constructed for Ex. 13 and 
compute the area of the whole 
surface. Find the volume also. 

15. A rectangular parallele- 
piped may be constructed from 

a figure like Fig. 92. Compute the volume of the solid and the 
area of the surface. 

SUMMARY 

150. This chapter has taught the meaning of the follow- 
ing words and phrases : a solid, surface of a solid, volume 
of a solid, unit of volume, cube, parallelepiped, rectangular 
parallelepiped, right parallelepiped, oblique parallelepiped, 
triangular pyramid, exponent, base, power, tetrahedron. 



FIG. 92. How TO CONSTRUCT A 
RECTANGULAR PARALLELEPIPED 



110 GENERAL MATHEMATICS 

151. The volume of a solid is determined by applying 
the unit cube to see how many times it is contained in the 
solid. The process is essentially comparison. The unit cube 
is a cube each of whose edges is one unit long. 

152. The following formulas have been used : 

v = Iwh, 
v=s 3 . 

153. The product of factors having a common base 
equals a number whose base is the same as the factors 
and whose exponent is the sum of the exponents of the 
factors. 

154. The quotient obtained by dividing a power by 
another power having the same base is a number whose 
base is the common base of the given powers and whose 
exponent is obtained by subtracting the exponent of the 
divisor from the exponent of the dividend. 



CHAPTER VI 

THE EQUATION APPLIED TO FUNDAMENTAL 
ANGLE RELATIONS 

155. Fundamental angle relations. In Chapter III we 
discussed the different kinds of angles and the methods of 
constructing them. In this chapter we shall study some 
of the fundamental relations between angles and see how 
the equation is applied to them. 

156. Relation of exterior sides of supplementary adjacent 
angles. Draw two adjacent angles of 64 and 116, of 75 
and 105, of 157 and 23. What is the sum of each pair? 
What is the relation of the exterior sides of each pair ? 




E 



FIG. 93 



157. Important geometric relation. The preceding article 
illustrates the following geometric relation : If the sum of 
two adjacent angles is a straight angle, their exterior sides 
form a straight line. 

ill 



112 GENEKAL MATHEMATICS 

EXERCISES 

1. Show that the geometric relation stated in Art. 157 agrees 
with the definition of a straight angle (Art. 61). 

2. In Fig. 93 read the number of degrees in angles XOA, 
XOB, XOC, XOY, XOD, XOE. 

3. What is the sum of /.XOA and Z.AOB? 

4. Express /.XOD as the sum of two angles. 

5. Express Z.AOB as the difference of two angles. 

6. Express /.XOE as the sum of three angles. 

158. Sum of all the angles about a point on one side 
of a straight line. Draw a line AB and choose a point 
P on it. Draw lines Pit, 

PS, and PT as shown in 
Fig. 94 and find the sum 
of the four angles formed. 
Estimate first and then 
measure with the pro- 
tractor. What seems to 
be the correct sum ? Ex- 
press the sum of the angles x, y, z, and w by means of 
an equation. Give a word statement for the equation. 

159. Important geometric relation. Art. 158 illustrates 
the truth of the geometric relation that the sum of all the 
angles about a point on one side of a straight line is a straight 
angle (180). 

EXERCISES 

1. Find the value of x and the size of each angle in Fig. 95. 

2. In the following examples each expression represents one 
of the angles into which all the angular space about a point 
on one side of a straight line has been divided. Write an 
equation expressing the sum of all the angles, solve for x, 




FUNDAMENTAL ANGLE RELATIONS 



113 



and find the size of each angle in degrees. Draw figures to 
illustrate your results in the first three examples. 

(a) x, 3 a-, 5 x + 9. 

(b) 3x, 4z-10, 33-2*, Wx + 7. 

(c) 2 x + 18}, 5 x + 91, 8J- + a-. 

(d) 2 (x + 5), 3 x + 24, 2 (35 + x). 

(e) 5.83 x, 3,94 a-, 1.27 a- + 11.55, 138.45 - 8.04 x. 





FIG. 95 



FIG. 96 



160. Sum of all the angles about a point in a plane. If 
we choose a point in a plane, as P in Fig. 96, and from this 
point draw four lines so as to make four angles, we can 
measure these angles and thus determine the sum of all 
the angles in a plane about a point. 



EXERCISES 

1. Measure the angles in Fig. 96 and write an equation 
expressing their sum. 

2. What is another way 
to show that the sum of the 
angles that exactly fill a plane 
about a point is two straight 
angles (360)? See the defi- 
nition of a perigon (Art. 61). 

3. Find the value of x and the size of each angle in Fig. 97. 




3Z+80 
FIG. 97 



114 GENERAL MATHKMATICS 

4. The expressions in the following examples represent the: 
angles into which the angular space about a point in a plane 
has been divided. Find the size of each angle. 

(a) 3 x, x, 2 x + 35, 125 - at. 

(b) 2x, 72 + 3x, 4* - 10, 118. 

(c) 10 x + 20-J-, 35| - x, 8 x + 49. 

(d) 5 a-, 3x + 27f,*7z - 20, 9a-. + 112j. 

(e) x + 1, 7 (a; + 1), 3 (35 + x), 2 x + 169. 

(f) 3x, 117 + 15 a-, 9 a; -27. 

(g) 14 x + 48, 28 x + 106f , 133^ - 6 x. 

The first two exercises in this article show that the sum 
of all the angles about a point in a j)lafie is 360. 

161. Left side of an angle; right side of an angle. If in 
Fig. 98 we imagine ourselves standing at the vertex of 
/.ABC and looking off over the angular space, say in the 
direction BD, then the side BC Q 

is called the left side of the angle 

(because it lies on our left), and - a* - 

the side BA is called the right s\f&^~~' 

side of the angle. ^-''" 

B Right A 

162. Notation. In lettering an- 

, , fi . . . f . FIG. 98 

gles and figures it is often desir- 
able to denote angles or lines that have certain characteristic 
likenesses by the same letter so as to identify them more 
easily. It is clear that to use I for the left side of one 
angle and the same I for the left side of another angle 
in the same discussion might be misleading. In order to 
be clear, therefore, we let ^ stand for the left side of one 
angle, Z 2 stand for the left side of a second angle, and 
1 3 stand for the left side of a third angle, etc. Then the 
three sides would be read " I sub-one," " I sub-two," " I sub- 
three," etc. 



FUNDAMENTAL ANGLE RELATIONS 



115 




li 



FIG. 



163. Important geometric relation. Two angles, a; 1 and 
x v in Fig. 99, are drawn so that their sides are parallel left 
to left and right to 

right. How do they 
seem to compare in 
size ? Check your 
estimate by measur- 
ing with a protractor. 
Give an argument 
showing that x-^ = x 
This article shows 

that if two angles have their sides parallel left to left and right 
to right, the angles are equal. 

EXERCISE 

Draw freehand two obtuse angles so that their sides will 
look parallel left to left and right to right. (The angles should 
be approximately equal. Are they ?) 

HINT. Take -two points for vertices and in each case imagine 
yourself standing at the point. Draw the left sides to your left and 
the right sides to your right. Assume the drawing correct and prove 
the angles equal. 

164. Important geometric relation. Two angles, x and 
y, in Fig. 100, have been drawn so that their sides are par- 
allel left to right and right to left. What relation seems 
to exist between them ? Meas- 
ure each with a protractor. 

Give an argument showing 
that z + # = 180. 

This article shows that if 
two angles have their sides par- 
allel left to right and right to 
left, their sum is a straight angle. F IG . 100 




1 1 




116 GENERAL MATHEMATICS 

EXERCISE 

Practice drawing freehand a pair of angles whose sides are 
parallel according to the conditions in the theorem of Art. 164. 
Is the sum approximately 180 ? 

165. Supplementary angles ; supplement. Two angles 
whose sum is equal to a straight angle (180) are said 
to be supplementary angles. Each 
angle is called the supplement of 
the other. 




166. Supplementary adjacent F 101 

angles. Place two supplementary 

angles adjacent to each other as in Fig. 101. Angles so 
placed are called supplementary adjacent angles. 

EXERCISES 

1. In Fig. 101 what is the angle whose supplement is Zee? 

2. In Fig. 102 are several angles, some pairs of which are 
supplementary. Make tracings of these angles on paper and 
by placing them adjacent decide which pairs are supplementary. 



FIG. 102 

3. State whether the following pairs of angles are sup- 
plementary : 40 and 140 ; 30 and 150 ; 35 and 135 ; 55 
and 135. 

4. How many degrees are there in the supplement of an 

angle of 30 ? of 90 ? of 150 ? of x ? 

2 s 

5. What is the supplement of y ? of z ? of 3 w ? of ? 

o 



117 

6. Write the equation which expresses the fact that y and 
130 are supplementary and solve for the value of y. 

7. Write equations that will show that each of the follow- 
ing pairs of angles are supplementary : 

(a) y and 80. (d) 30 and y + 40. 

(.b) 90 and z. (e) 3 y + 5 and 12 y - 4. 

(c) x and y. (f) f x and lx + 75 J. 

8. Two supplementary angles have the values 2x + 25 
and x + 4. Find x and the size of each angle. 

9. What is the size of each of two supplementary angles 
if one is 76 larger than the other ? 

10. One of two supplementary angles is 33 smaller than the 
other. Find the number of degrees in each. 

11.. What is the number of degrees in each of two supple- 
mentary angles whose difference is 95 ? 

12. Find the value of x and the angles in the following 
supplementary pairs : 

(a) x and 6 x. 

(b) 2 x and 3 x + 2. 

(c) 4 x and 6 x. 

(d) 2 aj + 5 and 7 x - 8. 

13. Write the following expressions in algebraic language : 

(a) Twice an angle y. 

(b) Four times an angle, plus 17. 

(c) 23 added to double an angle. 

(d) Seven times an angle, minus 14. 

(e) 45 less than an angle. 

(f ) 52 subtracted from four times an angle. 

(g) Twice the sum. of an angle and 10. 
(h) One half the difference of 22 and x. 

14. If an angle is added to one half its supplement, the 
sum is 100. Find the supplementary angles. 



118 GENERAL MATHEMATICS 

15. If an angle is increased by 5 and if one fourth of its 
supplement is increased by 25, the sum of the angles thus 
obtained is 90. Find the supplementary angles. 

16. Construct the supplement of a given angle. 

17. Find the size of each of the following adjacent pairs of 
supplementary angles : 



(a) 0llft : +i (d) -60, 130- *. 



(b) | x + 32, 88 - \x. (e) 2 (* + 10), ^ 

(c) | + 150, I - 10. (f) 65 + 2 -f, 92 + ^. 

167. Construction problem. To construct the supplements 
of two equal angles. 

Construction. Let a; and 
y be the given angles. 
Construct Z z, the supple- 

ment of Zz, adjacent to z /x w /y 

it (Fig. 103). In the same 
manner construct Zw-, the Fm - 103 ' HOWTO CONSTRUCT THE SCPPLK- 

, . MENTS OF T\VO GlVEX AHOLES 

supplement of Z.y. 

Compare the supplements of Z x and Z y and show that 

/-Z=/-W. 

This article shows that the Supplements of equal angles 
are equal. 

EXERCISES 

1. Prove the preceding fact by an algebraic method. 

HINT. In Fig.103 prove that if Zx + Zz =180 and Z.y + Zro =180, 
then Zz = Zw. 

2. Are supplements of the same angle equal ? Why ? 



FUNDAMENTAL ANGLE RELATIONS 



119 



3. Show that the bisectors of two supplementary adjacent 
angles are perpendicular to each other ; for example, in Fig. 104 
show that Z x + Z. y = 90. 

4. In Fig. 104, if Z. BOD = 60 
and Z. AOD = 120, find the size 
of Z. x and Z. y. 

*5. The following examples 
furnish a review. In each case 
solve for the value of the un- 
known, and check. 




() Y 

(b) \y 



OD 



(e) 



t ^ 

(g) 






. " 19 

h < = 12. 



16 + -* = 3. 
o 



168. Complementary angles. If the 
sum of two angles is a right angle, the 
two angles are called complementary 
angles. Each angle is called the com- 
plement of the other. Thus, in Fig. 105 
Z.x is the complement of Z y. 



FIG. 105 



EXERCISES 

1. What is the complement of 30 ? of 60 ? 

2. Are 23 and 57 complementary ? 32 and 58 ? 

3. Draw two complementary angles of 40 and 50 and 
place them adjacent. Check the construction. 

4. What is the relation existing between the exterior sides 
of two adjacent complementary angles ? 



120 



GENERAL MATHEMATICS 



5. In Fig. 106 decide by means of tracing paper which 
pairs of angles seem to be complementary. 

6. What are the complements of the following angles : 
20? 50? 12? 48|? x? 3y? ^? 

7. 40 is the complement of y. How many degrees does 
y represent ? 

8. Write the equation which says in algebraic language 
that x and 50 are complementary and solve for the value of x. 





FIG. 106 

9. In the equation x + y = 90 is there more than one 
possible pair of values of x and y ? Explain. 

10. Write equations that will express the fact that the 
following pairs of angles are complementary : 

(a) x and 40. (c) x + 25 and x - 30. 

(b) 35 and y. (d) 2 x - 3 and 3 x + 8. 

11. Write the following expressions in algebraic language: 

(a) The sum of angle x and angle y. 

(b) Four times an angle, increased by 15. 

(c) 85 diminished by two times an angle. 

(d) Five times the sum of an angle and 13. 

(e) Three times the difference between an angle and 12. 

(f) Four times an angle, minus 6. 



FUNDAMENTAL ANGLE RELATIONS 121 



12. Find two complementary angles such that one is 32 
larger than the other. 

13. Find two complementary angles such that one is 41 
smaller than the other. 

14. Find the number of degrees in the angle x if it is the 
complement of 3 x ; of 5 x ; of 8^ x - 

15. Find the number of degrees in the angle x if it is 
the complement of twice itself; of five times itself; of one 
third itself. 

16. Construct the complement of a given acute angle. 

169. Construction problem. To construct the comple- 
ments of two equal acute angles. 

Construction. Let Zx and /.y be the given angles. Draw the 
complement of x adjacent to it (Fig. 107). Do the same for y. 




FlG. 107. HOW TO CONSTRUCT THE COMPLEMENTS OF. TWO GlVEN ANGLES 

Compare the complements of Ax and y and show that 
Z0 = /.w. 

This construction problem shows that complements of 
equal angles are equal. 

EXERCISES 

1. Prove the preceding relation by an algebraic method. 

2. Does it follow that complements of the same angle are 
equal? Why? 



122 



GENERAL MATHEMATICS 



170. Vertical angles. Dra\v two in- 
tersecting straight lines AB and CD as 
in Fig. 108. The angles x and z are 
called vertical, or opposite, angles. Note 
that vertical angles have a common 
vertex and that their sides lie in the 
same straight line but in opposite direc- 
tions. Thus, vertical angles are angles 
which have a common vertex and their 
sides Jying in the same straight line -A D 

but in opposite directions. Are w and FIG. 108. VERTICAL 
y in Fig. 109 vertical angles? 




ANGLES 



EXERCISES 
(Exs. 1-6 refer to Fig. 108) 

1. Make a tracing of /- x and /- z and compare them as to size. 

2. Check your estimate in Ex. 1 by measuring the two angles 
with a protractor. 

3. What is the sum of /- x and /. y ? of Z. z and /- y ? 

4. Show that x + y = z -j- y. 

5. How does Ex. 4 help in obtaining the relation between 
x and z ? What is this relation ? 

6. Show that y + x = x -+- w and from this that y = w. 

The six exercises above show that if two lines intersect, 
the vertical angles are equal. 

171. Value of mathematical thinking. The preceding 
relation between vertical angles is of course so easily seen 
that in most cases the truth would be granted even with- 
out measuring the angles involved. However, the discus- 
sion in Exs. 3-6 above is another simple illustration of 



123 

the power of mathematical thinking which makes the dis- 
covery of new truths rest finally on nonmeasurement, that 
is to say, on an intellectual basis. This type of thinking 
will be used to an increasing extent in subsequent work. 

EXERCISES 

1. Upon what does the proof (Exs. 3-6, Art. 170) of 
the geometric relation concerning vertical angles rest ? 

2. Find x and the size of 

each angle in Fig. 109. ~^_^__ o 

First method. Since vertical 
angles are equal, 




Then 3r + 4 = 2ar + 10. 

Subtracting 4 from each member, 

3x = 2x + 6. 

Subtracting 2 x from each member, 

* = . 

Substituting 6 for x, 3 x + 4 = 3 6 + 4 = 22, 
2 x + 10 = 2 . 6 + 10 = 22, 
9 x + 104 = 9 . 6 + 104 = 158 (for Z. BOC), 
and since vertical angles are equal, 



Check. 22 + 22 + 158 + 158 = 360. 

Second method. By definition of supplementary angles, 

8* + 4 + 9* + 104 =180. 
Solving, . r = 6. 

The remainder of the work is the same as that of the first method. 



124 GENERAL MATHEMATICS 

3. Find the values of the unknowns and each of the follow- 
ing vertical angles made by two intersecting straight lines : 

(a) 3 x + 15 and 5x 5. (f) -J x + $ x and f x + 55. 

(b) , + 105 and IS* + 15. * 5, 

(c) cc. 10 and 2 a; 160. 4 2 

(d) + 181 and + 21. (h) + and + 18. 



Q ~ 8 T ^ -> 9 T 

(e)|*-8andj + 12. (i) ^ - - ^ and ^ + 11 ]. 

172. Alternate-interior angles. In Fig. 110 the angles 
x and y, formed by the lines AB, CD, and the transversal 
EF, are called alternate-interior angles (on alternate sides 
of EF and interior with respect to AB and CD~). 




FIG. 110 E FIG. Ill 



EXERCISES 

(Exs. 1-4 refer to Fig. Ill) 

1 . Measure and compare Z x and Z //. 

2. The lines AB and CD and FE are drawn so that Z.x=Z.y. 
What seems to be the relation between the lines AB and CD? 

3. Show that if Z x = Z y, then Z y = Z z. 

4. Show that 4-^ is parallel to CD (see the definition for 
parallel lines, Art. 87). 



FUNDAMENTAL ANGLE RELATIONS 



125 



Exercises 1-4 show that if the alternate-interior angles 
formed by two lines and a transversal are equal, the lines are 
parallel. 

The proof may take the following brief form: 

Proof. In Fig. Ill, Z x = Z y (given). Z x = Z z (vertical angles are 
equal). Then Z?/ = Z.z (things equal to the same thing are equal to 
each other). Therefore AB II CD (by definition of II lines, Art. 87). 

EXERCISE 

In Fig. 112 construct a line through P parallel to the line 
AB by making an alternate-interior angle equal to /.x. Show 
why the lines are parallel. F 

A \ B 





FIG. 112 

173. Interior angles on the same side of the transversal. 
In Fig. 113 angles x and y are called interior angles on the 
same side of the transversal. 

EXERCISES 

1. Measure angles x and y in Fig. 113 and find their sum. 

2. In Fig. 114 the lines are drawn so that /Lx + Z.y = 180. 



What seems to be the relation 
between AB and CD? 

3. Prove that if the interior 
angles on the same side of a, 
transversal between two par- 
allel lines are supplementary, 
the lines are parallel. 



'\- 




-D 



FIG. 114 




126 GENERAL MATHEMATICS 

4. In Fig. 115 select all the pairs of corresponding angles, 
alternate-interior angles, and in- F 

terior angles on the same side of / 

x /?/ 
the transversal. ./' B 

174. Important theorems relat- 
ing to parallel lines. The follow- c y^ D 

ing exercises include theorems / 

which supplement the work of 'E 

Arts. 172 and 173. FIG. 115 

EXERCISES 

1. Show by reference to the definition of parallel lines in 
Art. 87 that if two parallel lines are cut by a transversal, the 
corresponding angles are equal. 

2. Show that if two parallel lines are cut by a transversal, 
the alternate-interior angles are equal. 

3. Show that if two parallel lines are cut by a transversal, 
the interior angles on the same side of the transversal are 
supplementary. 

4. Two parallel lines are cut by a transversal so as to 
form angles as shown in Fig. 116. Find x and the size of 
all the eight angles in the figure. 

V Ab <K'^ 





\ X 

FIG. 116 FIG. 117 

5. Find x and all the eight angles in Fig. 117. 

6. Draw two parallel lines and a transversal. Select all the 
equal pairs of angles ; all the supplementary pairs. 



FUNDAMENTAL ANGLE RELATIONS 1.27. 

175. Outline of angle pairs formed by two lines cut by 
a transversal. When two lines are cut by a transversal, 

as in Fig. 118, 

"a and e~] 

the angles of the b and / 

7 IT?- are called corresponding angles ; 
angle pairs a and h 

c and g\ 

angles c, d, e, f are called interior angles ; 
angles a, 5, #, h are called exterior angles ; 

the angles of the f d and e 1 are called interior angles on the 
angle parrs j c and /j same side of. the transversal ; 

*:',, f , , J on opposite sides of the trans- 
the angles of the \ a and / 

, > versal are called alternate- 
angle pairs c and e\ 

J interior angles ; 

, j "] on opposite sides of the trans- 

the angles ot the b and h\ , . , 

^ , Y versal are called alternate- 

angle pairs a and a \ 

[ J exterior angles. 







FIG. 119 
The student should remember 

(a) that corresponding angles are equal, 

(b) that alternate-interior angles are equal, 

(c) that alternate-exterior angles are equal, 

(d) that interior angles on the same side of 

the transversal are supplementary, 
only when the lines cut by the transversal are parallel (Fig. 119). 



128 GENERAL MATHEMATICS 

SUMMARY 

176. This chapter has taught the meaning of the follow- 
ing words and phrases: left side of an angle, right side 
of an angle^ parallel right to right and left to left, par- 
allel right to left and left to right, supplementary angles, 
supplement, supplementary-adjacent angles, complementary 
angles, complement, vertical angles, alternate-interior angles, 
interior angles on the same side of the transversal. 

177. The following fundamental constructions have been 
presented : 

1. How to construct the supplement of a given angle. 

2. How to construct the supplements of two equal given 
angles. 

3. How to construct the complement of -a given angle. 

4. How to construct the complements of two equal 
angles. 

5. A new method of drawing parallel lines. 

6. How to form vertical angles. 

178. This chapter has discussed the following funda- 
mental geometric relations: 

1. If the sum of two adjacent angles is a straight angle, 
their exterior sides form a straight line. 

2. The sum of all the angles about a point on one side 
of a straight line is a straight angle (180). 

3. The sum of all the angles in a plane about a point 
is two straight angles (360). 

4. If two angles have their sides parallel left to left and 
right to right, the angles are equal. 

5. If two angles have their sides parallel left to right 
and right to left, the angles are supplementary. 

6. Supplements of the same angle or of equal angles 
are equal. 



FUNDAMENTAL ANGLE RELATIONS 129 

7. Complements of the same angle or of equal angles 
are equal. 

8. If two lines intersect, the vertical angles formed 
are equal. 

9. Two lines cut by a transversal are parallel 

(a) if the corresponding angles are equal ; 

(b) if the alternate-interior angles are equal; 

(c) if the interior angles on the same side of the trans- 
versal are supplementary. 

10. If two parallel lines are cut by a transversal, then 

(a) the corresponding angles are equal ; 

(b) the alternate-interior angles are equal ; 

(c) the interior angles on the same side of the trans- 
versal are supplementary. 



CHAPTER VII 



THE EQUATION APPLIED TO THE TRIANGLE 

179. Notation for triangles. It is customary to denote 
the three points of intersection of the sides of a triangle 
by capital letters and the three sides which He opposite 
these respective sides with the corresponding small letters. 
Thus, in Fig. 120 we denote 

the points of intersection 
of the sides (the vertices) 
by A, B, and C, and the 
sides opposite by a, 6, and c. 
The sides may also be 

read BC, AC, arid AB. The symbol for " triangle" is a small 
triangle (A). The expression A ABC is read " triangle ABC." 
The three angles shown in Fig. 120 are called interior angles. 

180. Measuring the interior angles of a triangle. We shall 
now consider some of the methods of measuring the interior 
angles of a triangle. 

EXERCISES 




ANGLE 


TRIANGLE ABC 
No. OF DEGREES 


TRIANGLE DEF 
No. OF DEGREES 


TRIANGLE GHI 
No. OF DEGREES 


Estimated 


Measured 


Estimated 


Measured 


Estimated 


Measured 


X 

y 

z 














Sum 















130 



131 



1. Fill in the table on the preceding page with reference 
to the triangles ABC, DEF, and GHI (Fig. 121). 






FIG. 121 

2. Draw a triangle on paper (Fig. 122). Cut it out and 
tear off the corners as shown in Fig. 123. Then place the 
three angles adjacent as shown. What 
seems to be the sum of the three angles 
of the triangle ? Test your answer with 
a straightedge. 

181. Theorem. The results of Exs. 1 FIG. 122 
and 2, above, illustrate the geometric 

relation that the sum of the interior 
angles of a triangle is a straight angle 
(180). The statement "The sum 
of the interior angles of a triangle is 
a straight angle" can be proved by more advanced geo- 
metric methods. Such a statement of a geometric relation 
to be proved is called a theorem. 

182. More advanced methods of 
proof for the preceding theorem. The 
truth of the theorem, that the sum of 
the interior angles of a triangle is Ji 
180 may be illustrated as follows: 
Draw a triangle as in Fig. 124. Place 

a pencil at A as indicated in the figure, noting the direction 
in which it points. Rotate the pencil through angle A as 



FIG. 123 




FIG. 124 



132 



GENERAL .M ATM K.MATK 'S 



sliown by the arrowhead. Then slide it along AB to the 
position indicated in the figure. Rotate the pencil next 
through angle B as indicated and slide it along BC to 
the position sliown. Then rotate the pencil through angle 
C to the last position shown. This rotation through 
angles .1. /;, and C leaves the point of the pencil in what 
position in respect to its original position ? What part of 
a complete turn has it made ? Through how many right 
angles has it turned ? Through 
how many straight angles ? 
Through how many degrees? 

The theorem that ' : the sum 
of the interior angles of a tri- 
angle is 180 " may be proved 
as follows : 

(oven triangle AB<'(Y\%. 12o), to prove that Z.I + Z/> + ZC' = 180. 

Proof 




STATE.MK.NT> 



REASONS 



Draw 



Then 



And 



Z if - Z B. 



But Z ./.; 4- Z // + A : = 180 



.-. z.i 



Because corresponding angles 
formed by two parallel lines cut 
by a transversal are equal. 

Because alternate-interior an- 
gles formed by two parallel lines 
cut by a transversal are equal. 

Because the sum of all the 
angles about a ] >oiut in a plane on 
one side of a straight line is 180. 

By substituting Z.A for Zr. 
i Z B for Z if. and Z C for Z r. 



This is n more formal proof of the theorem, inasmuch as it 
is independent of measurement. Write an equation which will 
express the number of degrees in the sum of the angles of 
a triangle. 



. EQUATION APPLIED TO THE TRIANGLE 133 

This equation is a very useful one, as it enables us to 
find one angle of a triangle when the other two are known. 
Thus, if we know that two angles of a triangle are 50 
and 70, we know that 60 is the third angle. This is of 
great practical value to the surveyor, who is thus enabled 
to know the size of all three angles of a triangle by 
measuring only two directly. 

HISTORICAL XOTK. Thales (040 B.C. - about o50 B.C.), the founder 
of the earliest Greek school of mathematics, is supposed to have known 
that the sum of the angles of a triangle is two right triangles. 
Someone has suggested that this knowledge concerning the sum 
of the angles of a triangle may have been experimentally demon- 
strated by the shape of the tiles used in paving floors in Th-ales' 
day. What has been regarded as the most remarkable geometrical 
advancement of Thales was the proof of a theorem which depended 
upon the knowledge that the sum of the angles of a triangle is two 
right angles. It is related that when Thales had succeeded in proving 
the theorem, he sacrificed an ox to the immortal gods. The large 
number of stories told about Thales indicates that he must have 
been a man of remarkable influence and shrewdness both in science 
and in business. Thus, we read that at one time he cornered the olive 
market and that at another time he was employed as engineer to 
direct a river so that a ford might be constructed. The following 
story is told illustrative of his shrewdness : 

It is said that once when transporting some salt which was loaded 
on mules one of the animals, slipping in a stream, got its load wet and so 
caused some of the salt to be dissolved. Finding its burden thus light- 
ened, it rolled over at the next ford to which it came ; to break it of 
this trick, Thales loaded it with rags and sponges, which, by absorbing 
the water, made the load heavier and soon effectually cured it of its 
troublesome habit. 1 

183. Problems involving the theorem " The sum of the 
interior angles of a triangle is a straight angle." In 
the problems that follow the pupil will need to apply the 
theorem proved in the preceding article. 

1 Ball, "A Short Account of the History of Mathematics," p. 14. 



134 GENERAL MATHEMATICS 

EXERCISES 

In the following problems 

(a) Draw freehand the triangle. 

(b) Denote the angles properly as given. 

(c) Using the theorem of Art. 182, write down the equation 
representing the conditions of the problem. 

(d) Solve the equation and find the value of each angle. 

(e) Check your solution" by the conditions of the problem. 

1. The angles of a triangle are x, 2x, and 3x. Find x and 
the number of degrees in each angle. 

2. The first angle of a triangle is twice the second, and the 
third is three times the first. Find the number of degrees in 
each angle. 

3. If the three angles of a triangle are equal, what is the 
size of each ? 

4. If two angles of a triangle are each equal to 30, what 
is the value of the third angle ? 

5. One angle of a triangle is 25. The second angle is 55 
larger than the third. How large is each angle ? 

6. The first angle of a triangle is four times the second, 
and the third is one half the first. Find e.ach angle. 

7. Find the angles of a triangle if the first is one half 
of the second, and the third is one third of the first. 

8. The first angle of a triangle is two fifths as large as 
another. The third is four times as large as the first. How 
large is each angle ? 

9. Find the angles of a triangle if the first angle is 16 more 
than the second, and the third is 14 more than the second. 

10. Find the angles of a triangle if the difference between 
two angles is 15, and the third angle is 43.. 

11. The first angle of a triangle is 30 more than the second, 
and the third is two times the first. Find the angles. 



EQUATION APPLIED TO THE TRIANGLE 135 

12. Find the angles of a triangle if the first angle is twice 
the second, and the third is 15 less than two times the first. 

13. The angles of a triangle are to each other as 1, 2, 3. 
What is the size of each ? 

HINT. Let x = the first, 2 x the second, and 3 x the third. 

14. Find the angles of a triangle if the first is 2^ times the 
second increased by 10, and the third is one fourth of the 
second. 

15. In a triangle one angle is a right angle; the other two 

%K> 

angles are represented by x and - respectively. Find each angle. 

16. How many right angles may a triangle have ? How 
many obtuse angles ? How many acute angles at most ? How 
many acute angles at least ? 

17. Two angles x and y of one triangle are equal respec- 
tively to two angles m and n of another triangle. Show that 
the third angle of the first triangle equals the third angle of 
the second triangle. 

184. Theorem. By solving Ex. 17 we obtain the theorem 
If two angles of one triangle are equal respectively to two 
angles* of another triangle, the third angle of the first is equal 
to the third angle of the second. 

185. Right triangle. If one angle of a triangle is a right 
angle, the triangle is called a right triangle. The symbol 
for " right triangle " is rt. A. 

EXERCISES 

1. Show that the sum of the acute angles of a right triangle 
is equal to a right angle. 

2. Find the values of the acute angles of a right triangle if 
one angle is two times the other ; if one is 5 more than three 
times the other. 



130 



GENEKAL MATHEMATICS 



/' /' 

3. The acute angles of a right triangle are ^ and r 



Find 



and the number of degrees in each angle. 

4. Draw a right triangle on cardboard so that the two acute 
angles of the triangle will contain 30 and 60 respectively. 
Use the protractor. 

HINT. First draw a right angle. Then at any convenient point 
in one side of the right angle construct an angle of 60 and produce 
its side till a triangle is formed. Why does the third angle equal.30? 

*5. Cut out the cardboard triangle made in Ex. 4 and tell 
what angles may be constructed by its use without a protractor 
or tracing paper. 

6. Draw on cardboard a right triangle whose acute angles 
are each equal to 45, cut it out, and show how it may be used 
to draw angles of 45 and 90 respectively. 

186. Wooden triangles. A wooden triangle is a triangle 
(usually a right one) made for convenience in drawing 
triangles on paper or on the blackboard (see Fig. 126). 
The acute angles are usually 60 and 30 

or 45 and 45. These 
wooden right triangles fur- 
nish a practical method of 
drawing a perpendicular to 
a line at a given point on 
that line. If no triangles 
of this kind can be had, 
a cardboard with two per- 
pendicular edges or a card- 
board right triangle will serve the purpose just as well. 

187. Set square. A set square is made up of a wooden 
triangle fastened to a straightedge so that it will slide 
along the straightedge (see Fig. 127). 





FIG. 126. WOODEN 
TRIANGLE 



FIG. 127. SET 
SQUARE 



EQUATION APPLIED TO THE TRIANGLE 137 

EXERCISES 

1. Construct a right triangle, using the method of Art. 80. 

2. The set square (Fig. 127) is a mechanical device for drawing 
parallel lines. Show how it may be used to draw parallel lines. 

3. Show that two lines perpendicular to the same line are 
parallel (see the definition of parallel lines, Art. 87). 

4. Show how a wooden triangle may be used to draw 
parallel lines. 

5. What are three ways of drawing parallel lines? 

188. Problems concerning the acute angles of a right 
triangle. The following problems will help the student to 
understand the relations concerning the acute angles of 
a right triangle. 

EXERCISES 

1. Draw a right triangle as in Fig. 128 and show that angles 
. I and B are complementary ; that is, show that /. A + /.B = 90. 

NOTE. In lettering a right triangle 
ABC the letter C is usually put at the 
vertex of the right angle. 

2. State a theorem concerning 
the acute angles of a right-angled 
triangle. 

3. If Z.I =Z B (Fig. 128), what 

is the size of each ? How do the sides about the right angle 
compare in length ? 

4. If /.A is twice as large as /.B (Fig. 128), what is the size 
of each ? 

*5. Using the method of Ex. 4, Art. 185, draw a right triangle 
whose acute angles are 30 and 60 respectively. Measure 
the side opposite the 30-degree angle. Measure the side oppo- 
site the 90-degree angle. (This side is called the hypotenuse.) 
Compare the two results obtained. 




138 



GENERAL MATHEMATICS 




6. To measure the distance AB across a swamp (Fig. 129), a 
man walks in the direction AD, so that /.BAD= 60, to a point 
C, where ^BCA = 90. If AC = 300 yd., 

what is the length of AB ? 

7. Find the number of degrees in each 
acute angle of a right triangle if one angle is 

(a) four times the other ; 

(b) three fourths of the other ; 

(c) two and a half times the other ; 

(d) 5" more than three times the other ; 

(e) 5 less than four times the other. 

*8. Practice drawing angles of 30, 45, 60, and 90 by 
using wooden or cardboard triangles. 

Ex. 5 illustrates the truth of the theorem In a right 
triangle whose acute angles are 30 and 60 the side opposite 
the 30 angle is one half the hypotenuse. This theorem will 
be proved formally later. It is very important because of 
its many practical applications in construction work and 
elsewhere. 

189. Isosceles triangle ; base angles. A triangle which 
has two equal sides is called an isosceles triangle. The 
angles opposite the equal sides are called the base angles of 
the isosceles triangle. 

EXERCISES 

1. Two equal acute angles of a right triangle are repre- 
sented by 2 x + 5 and 3 x 15. Find the size of each angle. 

2. Draw a right triangle ABC 
(Fig. 130). Draw a line from 
CAB; call the foot of the per- 
pendicular P. Show that the 
perpendicular (CP) divides the 

A ABC into two right triangles. FIG. 130 




EQUATION APPLIED TO THE TRIANGLE 139 

3. In Ex. 2 the angle x is the complement of what two 
angles ? What is the relation between these two angles ? 

\4\ In Ex. 2, /.y is the complement of two angles. Indicate 
them. Show that /. 



5. Draw freehand three isosceles triangles. 

190. Scalene triangle. A scalene triangle is a triangle 
no two of whose sides are equal. 

EXERCISES 

1. Draw freehand a scalene triangle. 

2. Do you think that a right triangle whose acute angles 
are 30 and 60 is a scalene triangle ? Support your answer. 

3. Draw an obtuse- 
angled scalene triangle. 

191. Exterior angles 
of a triangle. If the 
three sides of a tri- 
angle are extended, one 

at each vertex, as in '' 

Fig. 131, the angles thus FlG - 13L I^STRATING THE EXTERIOR 

ANGLES OF A TRIANGLE 
formed (x, y, and z) 

are called exterior angles of the triangle ABC. 

EXERCISES 

1. How many exterior angles can be drawn at each vertex 
of a triangle ? 

2. How many interior angles has a triangle ? How many 
exterior angles ? 

3. Draw a triangle and extend the sides as in Fig. 131. 
Measure the three exterior angles with a protractor. What is 
their sum ? 




14< 



GENERAL MATHEMATICS 




FIG. 132 



4. I>ra\\ another triangle and extend the sides as in. Fig. 131. 
Cut out the exterior angles (taking one at each vertex) with a 
pair of scissors and place them next to 

each other with their vert ices together. 
What does their sum seem to be'/ 

5. Find the sum of the three ex- 
terior angles x, //. and .-.- in Fig. 132 In- 
rotating a pencil as indicated by the 

arrowheads. 

6. Show that tin- mini nf flic e;rti'ri<n- 

niKjli-x i >f a triangle (taking one at each vertex) /\ -><>0 (two 
straight angles). 

HINT. How many degrees are in the sum x + w/? // + n ? z + r? 
(S,-,. Fig. l:;i.) 

Show that tlie sum 

(.' + m) + (// + n) + (z + ?) = 3 X 180 = 540. 
Then this fact may be expressed as follows : 

(x + ?/ + r) + (HI + n + r) = 540. 

But ( x + y + z -) = 180. ^ Why ? 

Therefore (m + n + r) = :5i;o . \Vliy? 

7. The three interior angles of a triangle are equal. Find 
the size of each interior and each exterior angle. 




FIG. 133 



8. Find the value of the interior and exterior angles in the 
triangle of Fig. 133. 



EQUATION APPLIED TO THE TRIANGLE 141 




FICJ. 134 



9. Show that the exterior angle x of the triangle AB(.' 
in Fig. 134 is equal to the sum of the two nonadjacent interior 
angles A and C. 

10. Using Fig. 135, in which 
BD II ,1 (7, prove that an ex- 
terior (inrjle of a triangle is / -\ \x 
('ijiinl f'> flif sifin of the two A 

nonadjficent interior angles. 

Note that two different methods are suggested by this figure. 

11. Prove Ex. 10 by drawing a line through C parallel to AB. 
HINT. Extend line A C. 

12. Draw a quadrilateral. Tear off the corners and place 
the interior angles 

next to each other 
by the method of 
Ex. 2, Art. 180. 
What does the sum 
of the interior an- 
gles seem to be ? 

13. Draw a quadrilateral as in Fig. 136. Draw the diagonal 
.4 C. This divides the quadrilateral into two triangles. What is 
the sum of the interior angles 

in each triangle ? What, then, 
is the sum of the interior 
angles of a quadrilateral ? 

14. Draw a quadrilateral as 
in Fig. 136. Produce each side 
(one at each vertex). What 
do you think is the sum of 
the exterior angles of the 

quadrilateral ? Check your estimate by measuring the angles. 

15. Find the angles of a quadrilateral in which each angle 
is 25 smaller than the consecutive angle. 





FIG. 136 



142 GENEEAL MATHEMATICS 

16. Prove that the consecutive angles of a parallelogram 
are supplementary ; that is, prove x + y = 180, in Fig. 137. 

'17. Prove that the 

opposite angles of a par- _Dl /<?___ 

allelogram are equal. 

HINT. In Fig. 137 show 
that Zx = Za = Zz. 

18. If one angle of 

a parallelogram is twice ' FIG. 137 

as large as a consecutive 

angle, what is the size of each angle in the parallelogram? 

19. The difference between two consecutive angles of a 
parallelogram is 30. Find the size of all four angles in the 
parallelogram. 

20. Show that the sum of the interior angles of a trapezoid 
is two straight angles (180). 

21. Prove that two pairs of 
consecutive angles of a trape- 
zoid are supplementary. (Use 
Fig. 138.) 

22. In Fig. 138, Z D is 40 Fl0 ' 138 

more than Z. A, and /.B is 96 less than Z. C. Find the number 
of degrees in each angle. 

192. The construction of triangles. We shall now proceed 
to study three constructions which require the putting 
together of angles and line segments into some required 
combination.. With a little practice the student will see 
that the processes are even simpler than the thinking 
involved in certain games for children which require the 
various combinations of geometric forms. 

These constructions are very important in all kinds of 
construction work ; for example, in shop work, mechanical 



EQUATION APPLIED TO THE TRIANGLE 143 

drawing, engineering, and surveying. The student should 
therefore master them. 

193. Construction problem. To construct a triangle when 
the three sides are given. 

Construction. Let the given sides be a, b, and c, as shown in Fig. 139. 

Draw a working line X Y, and lay off side c, lettering it AB. With 

A as a center and witk a radius equal to b construct an arc as shown. 




c 
FIG. 139. How TO CONSTRUCT A TRIANGLE WHEN THREE SIDES 

ARE GIVEN 

With B as a center and with a radius equal to a construct an arc 
intersecting the first. Call the point of intersection C. Then the 
triangle is constructed as required. 

EXERCISES 

v 1. Construct triangles with the following sides : 
- (a) a = 5 cm., b = 5 cm., c = 8 cm. 

(b) a = 1 cm., b = 8 cm., c = 4 cm. 

(c) a = 7 cm., b = 9 cm., c = 3 cm. 

2. Is it always possible to construct a triangle when three 
sides are given ? a 

3. Construct a triangle, 

using the sides given in | 1 

Fig. 140. 

. c 

4. Compare as to size and 

, , . , , FIG. 140 

shape the triangle drawn by 

you for Ex. 3 with those drawn by other pupils. (See if the 
triangles will fold over each other.) 



1 (JKNKKAL MATHEMATICS 

5. Make a wn.id.-n triangle by nailing three sticks together. 
Is it possible to change the shape of the triangle without break- 
ing a stick or removing the corner nails '/ 

6. A great deal of practical use is made of the fact that 
a triangle is a rigid figure ; for example, a rectangular wooden 
gate is usually divided into two triangles by means of a wooden 
diagonal so as to make the gate more stable (less apt to sag). 
Try to give other examples of the practical use that is made 

^of the stability of the triangular figure. 

7. Construct an isosceles triangle having given the base 
and one of the two equal sides. 

HINT. Use c in Fig. 140 as the base and use // twice: that is, in 
Uais case take a = />. 

8. Measure the base angles of the isosceles triangle drawn 
for Ex. 7. What appears to be the relation between the base 

.angles of an isosceles triangle '.' 

9. Make tracings of the base angles drawn for Ex 7 and 
attempt to fold one angle over the other. Do the two angles 
appear to represent the same amount of rotation ? 

10. Compare your results witli those obtained by your 
classmates. 

NOTE. Results obtained from Exs. 7-10 support the following 
theorem : Tin- buxe anglt-x of an isosceles triangle fire ef/>inl. 

11. Construct an equilateral triangle having given a side. 

12. Study the angles of an equilateral triangle by pairs in 
the manner suggested by Exs. 8-9. State 

the theorem discovered. 

13. To measure the distance AB 
(Fig. 141) we walk from B toward M 

so that Z/J = 50, until we reach C, a / 

point at which /.ACB = 50. What line 

must we measure to obtain A B ? Why ? FIG. 141 




EQUATION APPLIED TO THE T1UANULE 145 



14. At a point C (Fig. 142), 70 ft. from the foot of a 
pole AB, the angle ACB was found to be 45. How high is 
the pole ? 

*15. Walking along the bank of a river from A to C 
(Fig. 143), a surveyor measures at B the angle ABD, and 
walking 500 ft. further to C, he finds angle BCD to be one 
half of angle A BD. What is the distance BD ? 

D 




C A 

FIG. 142 



FIG. 143 



194. Construction problem. To construct a triangle, hav- 
ing given two sides and the angle included between the 
sides. 

Construction. Let the given sides be a and b and the given angle 
be Z C, as shown in Fig. 144. 



B 





FIG. 144. How TO CONSTRUCT A TRIANGLE WHEN Two SIDES AND THE 
INCLUDED ANGLE ARE GIVEN 

Draw a working line XY, and lay off CA equal to b. At C draw 
an angle equal to angle C by the method of 78. With C as a 
center and a radius equal to a lay off CB equal to a. Join B and A, 
and the triangle is constructed as required. 



146 



GENERAL MATHEMATICS 



EXERCISES 

1. Is the construction given in Art. 194 always possible ? 

2. Draw triangles with the following parts given :' 

(a) a = 3 cm., b = 4 cm., Z C = 47. 

(b) c = lin., 6 = 2i 

(c) J = ljin., c = lf 

3. Construct a triangle with the parts given in Fig. 145. 



Z.I = 112. 

= 87. 




FIG. 145 

4. Compare as to size and shape the triangle drawn for Ex. 3 
with those drawn by other students in your class. (Place one 
triangle over the other and see if they fit.) 

195. Construction problem. To construct a triangle when 
two angles and the side included between them are given. 





FIG. 146. How TO CONSTRUCT A TRIANGLE WHEN Two ANGLES AND THE 

SlDE INCLUDED BETWEEN THEM ARE GIVEN 



Construction. Let ZA and ZB be the given angles and line c be 
the given included side (Fig. 146). 



EQUATION APPLIED TO THE TRIANGLE 147 

Lay down a working line A'Fand lay off AB equal to linec on^it. 
At A constrict an angle equal to the given angled; at B construct 
an angle equal to the given angle B and produce the sides of those 
angles till they meet at C, as shown. Then the t\ABC is the 
required triangle. 

EXERCISES 

1. Draw triangles with the following parts given : 
(a) Z A = 30, Z B = 80, e = 4 cm. 



(b) Z C = 110, Z = 20, a = 2 in. 

2. Draw a triangle with the parts as given in Fig. 147. 

3. Is the construction of Ex. 2 always possible ? 




FIG. 147 

4. Compare as to size and shape the triangle drawn for Ex. 2 
with those drawn by other members of your class. (Fold them 
over each other and see if they fit.) 

SUMMARY 

196. This chapter has taught the meaning of the follow- 
ing words and phrases : right triangle, cardboard triangle, 
wooden triangle, set square, isosceles triangle, scalene 
triangle, interior angles of a triangle, exterior angles of 
a triangle, base angles. 

The following notations have been given : notation 
for the angles and sides of triangles, notation for right 
triangle (rt. A). 



148 

197. This chapter has presented methods of 

1. Finding the sum of the interior angles of a triangle. 
Finding the sum of the exterior angles of a triangle. 

3. Drawing right triangles by means of wooden or card- 
board triangles. 

4. Drawing parallel lines by means of the wooden tri- 
angle or the set square. 

198. This chapter has taught the pupil the following 
constructions : 

1. Given three sides of a triangle, to construct the triangle. 

2. Griven two sides and the included angle of a triangle, 
to construct the triangle. 

3. Given two angles and the included side of a triangle, 
to construct the triangle. 

199. The following theorems have been presented in 
this chapter: 

1. The sum of the interior angles of a triangle is a 
straight angle (180). 

2. The sum of the exterior angles of a triangle is two 
straight angles (360). 

3. If two angles of one triangle are equal respectively 
to two angles of another triangle, the third angle of the 
first triangle is equal to the third angle of the second. 

4. The acute angles of a right triangle are comple- 
mentary. 

5. In a right triangle whose acute angles are 30 and 
60 the side opposite the 30-degree angle is one half the 
hypotenuse. 

6. An exterior angle of a triangle is equal to the sum 
of the two nonadjacent interior angles. 

7. The sum of the interior angles of a quadrilateral is 
four right angles (360). 



8. The sum of the exterior angles of a quadrilateral 
is four right angles (360). 

9. The opposite angles of a parallelogram are equal. 

10. The consecutive angles of a parallelogram are 
supplementary. 

11. Two pairs of consecutive angles of a trapezoid are 
supplementary . 

12. The base angles of an isosceles triangle are equal. 

13. An equilateral triangle is equiangular (all angles 
equal). 



CHAPTER VIII 

POSITIVE AND NEGATIVE NUMBERS. ADDITION AND 
SUBTRACTION 

200. Clock game. Mary and Edith were playing with a 
toy clock. Each took her turn at spinning the hand. The 
object of the game was to guess the number on which the 
hand of the clock stopped. A correct guess counted five 
points. If a player missed a guess by more than three, 
she lost three points. If she came within three she either 
won or lo'st the number of points missed, according to 
whether she had guessed under or over the correct number. 
After five guesses they had the following scores : 

MARY Solution. The score as kept by the 

Won 2 players (the words are inserted) appeared 

Lost 1 as follows : 

Lost 3 
Won 2 
Won 1 

EDITH 
Lost 1 
Lost 2 
Won 5 
Won 2 
Lost 2 

Who won the 

game? Edith won the game, 2 to 1. 

150 



MARY 




EDITH 




First score 





First score 


CD 


Lost 


1 


Lost 


2 


Second score 


1 


Second score 


(D 


Lost 


3 


Won 


5 


Third score 


d) 


Third score 


2 


Won 


2 


Won 


2 


Fourth score 





Fourth score 


4 


Won 


1 


Lost 


2 


Final score 


1 


Final score 


2 



POSITIVE AND NEGATIVE NUMBERS 151 

201. Positive and negative numbers. Algebraic numbers. 
The adding of scores in many familiar games like the one 
cited above illustrates an extension of our idea of count- 
ing that will be found very useful in our further study of 
mathematics. It is important to notice that the players 
began at zero and counted their scores in both directions. 
Thus, Mary began with zero and won two points, and the 
2 she wrote meant to her that her score was two above 
zero. On the other hand, Edith had zero and lost one. 
In order to remember that she was " one in the hole " 
she wrote 1 within a circle. On the next turn she lost 
two more, and she continued to count away from zero 
by writing 3 within a circle. The same idea is shown in 
Mary's score. Her second score was 1. On the next turn 
she lost three. She subtracted 3 from 1. In doing so she 
counted backward over the zero point to two less than zero. 
In writing the scores it was necessary to indicate whether 
the number was above or below zero. 

We shall presently have numerous problems which 
involve pairs of numbers which possess opposite qualities, 
like those above. It is generally agreed to call num- 
bers greater than zero positive and those less than zero 
negative. Such numbers are called algebraic numbers. 

The opposite qualities involved are designated by the 
words " positive " and " negative." In the preceding game, 
numbers above zero are positive, whereas numbers below 
zero are negative. 

202. Use of signs. To designate whether a number is 
positive or negative we use the plus or the minus sign. 
Thus, 4- 4 means a positive 4 and 4 means a negative 4. 
The positive sign is not always written. When no sign pre- 
cedes a number the number is understood to be a positive 
number. Thus, 3 means + 3. 



152 

The following stock quotations from the Chicago Daily 
Tribune (March 24, 1917) illustrate a use of the plus and 

minus signs : 

CHICAGO STOCK EXCHANGE 





SALES 


HIGH 


Low 


CLOSE 


NET 


American Radiator . . . 


20 


295J 


295 


296 


-2 


Commonwealth Edison . . 


79 


136i 


136 J 


136 J: 


-1 


Diamond Match 


40 


81 


81 


81 





Swift & Co 


515 


114^ 


142f 


1441 


+ ~'- 


Peoples Gas 


397 


95" 


90^ 


91 


- 5 


Prest-o-Lite 


40 


1292 


129 


129 


-3 


Sears Roebuck 


325 


192 


190* 


19H 


~ <> 


Booth Fisheries 


20 


79 


79 


79 


+ T 















The last column shows the net gain or loss during the 
day; for example, American Radiator stock closed two 
points lower than on the preceding day, Swift & Co. 
gained 2|, Peoples Gas lost 5, etc. The man familiar 
with stock markets glances at the first column to see 
the extent of the sales and at the last column to see 
the specific gain or loss. To check the last column one 
would need the quotations for the preceding day. 

In order that we may see something of the importance 
of the extension of our number system by the preceding 
definitions, familiar examples of positive and negative 
numbers will be discussed. 

203. Geometric representation of positive numbers. Origin. 
We have learned in measuring a line segment that when a 
unit (say^[_^) is contained five times 'in another segment, 
the latter is five units long. (In general, if it is contained 
a times, the measured segment is a units long.) We may 
also say that the two segments represent the numbers 1 and 
o respectively. This suggests the following representation 



POSITIVE AND NEGATIVE NUMBERS 153 

of positive integers : On any line OX (Fig. 148) beginning 
at (called the origin of the scale) but unlimited in the 
direction toward A', we lay off line segments OA, AB, BC, 
CD, DE, etc., each one unit in length. We thus obtain a 
series of points on the o A B c D E F 

line, each of which I 1 1 \ 1 1 1 1 1 x 

corresponds to some 012345678 
positive integer, and FlG - 148 - GEOMETRIC REPRESENTATION OF 
T , r , POSITIVE NUMBERS 

vice versa. We also 

note that the line segment which connects the origin with 
some point (say JE") is the corresponding number of units 
in length. (Thus, OE = 5.) Hence the positive integers 
+ 1, -f 2, -f 3, + 4, 4- 5, etc. are represented by OA, OB, 
OC, OD, OE, etc. 

204. Geometric representation of negative numbers. Num- 
ber scale. If we prolong OX to the left from in the 
direction OX' (Fig. 149), we may lay off line segments 
in either of two opposite directions. Segments OC and 
OD' differ not only hi length but also in direction. This 
difference in direction is indicated by the use of the 

D' C' B 1 A' A B C D 

X r 1 1 1 1 1 1 1 1 1- 

-4-3-2-1 1 2 3 4 

FIG. 149. GEOMETRIC REPRESENTATION OF NEGATIVE NUMBERS 

signs + and . Thus, OC represents + 3, whereas OD' rep- 
resents 4. The integers of algebra may now be arranged 
in a series on this line called the number scale, beginning 
at and extending both to the right and to the left. 
In order to locate the points of the number scale (Fig. 149) 
we need not only an integer of arithmetic to determine 
how far the given point is from zero but also a sign of 
quality to indicate on which side of it is found. 




154 GENEKAL MATHEMATICS 

EXERCISES 

1. Locate the following points on the number scale : + 3, 
-3, -7, + 12, -8, +6. 

2. If we imagine that each of the players in the clock 
game (Art. 200) has a string or tape measure graduated to 
the number scale, with a ring on it, they could add their 
scores from time to time by sliding the ring back and forth. 

Indicate on the number scale how the sum of the following 
consecutive scores could be found : Began at 0, lost 1, lost 2, 
won 5, won 2, lost 1. 

3. Starting from the middle of the field in a certain football 
game, the ball shifted its position in yards during the first 
fifteen minutes of play as follows: .+ 45, 15, +11, 10, 
2, + 23. Where was the ball at the end of the last play ? 

205. Addition and further use of positive and negative 
numbers. The preceding exercises show that positive and 
negative numbers may be added by counting, the direc- 
tion (forward or backward) in which we count being 
determined by the sign (+ or ) of the numbers which 
we are adding. Thus, 

To add + 4 to + 5 on the number scale of Fig. 149, begin at + 5 
and count 4 to the right. 

To add 4 to + 5 begin at + 5 and count 4 to the left. 

To add 4 to 5 begin at 5 and count 4 to the left. 

To add + 4 to 5 begin at 5 and count 4 to the right. 

The results are as follows : 

+ 5 +(+4) = + 9; +5+(-4) = + l; -5 + (-4) = -9; 



+ 5 + ( 4) = -f 1 is read "positive 5 plus negative 4 equals 
positive 1." 

5 +(4) = 9 is read "negative 5 plus negative 4 equals 
negative 9." 



7 



POSITIVE AND NEGATIVE NUMBERS 155 

EXERCISES 

1. Give the sum in each of the following. Be prepared to 
interpret the result on the number scale. 

- (a) 3 +(+2). (g) 6+(-l). - (m) -2+(-5). 

(b) 4 +(- 3). (h) 6 +(- 3). ^(n) - 4 + 6. 

2. On a horizontal straight line, as X'OX in Fig. 150, con- 
sider the part OX as positive, and the part OX' as negative. 
Construct line segments correspond- 
ing to the following numbers : 2, 

+ 6, - 3, - 5, 0. 

3. Consider OF as positive and OY' x r 
as negative and construct segments on 

the line YOY' corresponding to 4, 
- 2, + 3, + 4, - 3, 0. 

4. A bicyclist starts from a certain 

point and rides 18 mi. due northward -p IG JCQ 

(4- 18 mi.), then 12 mi. due south- 
ward ( 12 mi.). How far is he from the starting point? 

5. How far and in what direction from the starting point is 
a traveler after going eastward (+) or westward ( ) as shown 
by these pairs of numbers : + 16 mi., then 3 mi.? 2 mi., then 
-f- 27 mi.? 16 mi., then + 16 mi.? + 100 mi., then 4- 52 mi.? 

6. Denoting latitude north of the equator by the plus sign 
and latitude south by the minus sign, give the meaning of the 
following latitudes : 4- 28, + 12, - 18, + 22, - 11. 

7. Would it be definite to say that longitude east of Green- 
wich is positive and west is negative ? What is the meaning 
of the following longitudes': 4- 42? + 142? - 75? - 3 ? 



156 GENERAL MATHEMATICS 

8. A vessel starting in latitude + 20 sails + 17, then G3, 
then + 42, then 1G. What is its latitude after all the 
sailings ? 

9. What) is the latitude of a ship starting in latitude 53 
after the following changes of latitude : + 12, - 15, + 28, 
- 7, + 18, - 22, + 61 ? 

206. Double meaning of plus and minus signs. Absolute 
value. The many illustrations of negative number show 
that there is a real need in the actual conditions of life 
for the extension of our number scale so as to include 
this kind of number. It must be clear-ly understood that 
a minus sign may now mean two entirely different things : 
(1) It may mean the process of subtraction, or (2) it may 
mean that the number is a negative number. In the latter 
case it denotes quality. Show that a similar statement is 
true for the plus sign. 

This double meaning does not often confuse us, since it is 
usually possible to decide from the context of the sentence 
which of these two meanings is intended. Sometimes a 
parenthesis is used to help make the meaning clear, thus 
4 + ( 3) means add a negative 3 to a positive 4. 

Sometimes we wish to focus attention merely on the 
number of units in a member regardless of sign. In that 
case we speak of the absolute value, or numerical value. 
Thus, the absolute value of either -f 4 or 4 is 4. 

207. Forces. In mechanics we speak of forces acting 
in opposite directions as positive and negative. Thus, a 
force acting upward is positive, one acting downward is 
negative. 



POSITIVE AND NEGATIVE NUMBERS 157 



EXERCISES 

1. Three boys are pulling a load on a sled, one with a force 
of 27 lb., another with a force of 56 lb., and the third with a 
force of 90 lb. With what force is the load 
being pulled ? 

2. Two small boys are pulling a small 
wagon along; one pulls with a force of 
23 lb., and the other pulls with a force of 
36 lb. A boy comes up behind and pulls 
with a force of 47 lb. in the opposite direc- 
tion from the others. What is the result ? 




FIG. 151 



3. An aeroplane that 
can fly 48.3 rni. an hour 
in still air is flying 
against a wind that 

retards it 19.6 mi. an hour. At what rate 

does the aeroplane fly ? 

4. A balloon which exerts an upward 
pull of 512 lb. has a 453-pound weight 
attached to it. What is the net upward 
or downward pull ? 

5. A toy balloon (Fig. 151) tends to pull 
upward with a force of 8 oz. What happens 
if we tie a 5-ounce weight to it ? an 8-ounce ? 

6. A boy can row a boat at the rate of 
4 mi. per hour. How fast can he go up 
a river flowing at the rate of 2^ mi. per 
hour ? How fast can he ride down the 
river ? How fast could he go up a stream 
flowing 5 mi. per hour ? 

208. The thermometer. The ther- 
mometer (Fig. 152) illustrates the idea 
of positive and negative numbers in two 



Z12- 



100- 



70- 



32- 



FIG. 152. THE TIIKK- 
SIOMETER ILLUSTRATES 
THE IDEA OF POSI- 
TIVE AND NEGATIVE 

NUMBEBS 



158 GENERAL MATHEMATICS 

ways. In the first place, the number scale is actually pro- 
duced through the zero, and degrees of temperature are 
read as positive (above zero) or negative (below zero). 
In the second place, the thermometer illustrates positive 
and negative motion discussed in the preceding article. 
Thus, when the mercury column rises, its change may be 
considered as positive, + 5 indicating in this case not a 
reading on the thermometer, as before, but a change (rise) 
in the reading ; similarly, 3 in this sense indicates a 
drop of 3 in the temperature from the previous reading. 

EXERCISES 

1. What is the lowest temperature you have ever seen 
recorded ? 

2. The top of the mercury column of a thermometer stands 
at at the beginning of an hour. The next hour it rises 5, 
and the next 3; what does the thermometer then read ? 

3. If the mercury stands at 0, rises 8, and then falls 5, 
what does the thermometer read ? 

4. Give the final reading in each case : 

A first reading of 10 followed by a rise of 2. 
A first reading of 10 followed by a fall of 12. 
A first reading of 20 followed by a fall of 18. 
A first reading of x followed by a rise of y. 
A first reading of x followed by a fall of y. 
A first reading of a followed by a rise of a. 
A first reading of a followed by a fall of a. 
A first reading of a followed by a fall of a. 

5. The reading at 6 P.M. was 7. What was the final read- 
ing if the following numbers express the hourly rise or 
fall : + 2, + 1, 0, - 3, - 3, - 2, - 2, - 1, - 1, - 3, 
1 4- 2 ? 



POSITIVE AND NEGATIVE NUMBERS 159 

6. Add the following changes to find the final reading, the 
first reading being : + 3, + 2, - 4, - 3, - 2, -f 3. 

7. The differences in readings of a thermometer that was 
read hourly from 5 A. M. until 5 P. M. were as follows : 3, 4, 
7, 10, 12, 9, 8, 5, 0, -22, -17, -12. How did 
the temperature at 5 P.M. compare with 'that at 5 A.M.? 
If the. temperature at 5 A.M. was -f- 20, make a table showing 
the temperature at each hour of the day. 

209. Positive and negative angles. By rotating line AB 
in a plane around A until it takes the position AC the 
angle BAG is formed (Fig. 153). 
By rotating AB in the opposite direc- 
tion angle BAC is formed. To dis- 
tinguish between these directions 
one angle may be denoted by the 
plus sign, and the other by the FlG 153 

minus sign. We agree to consider 

an angle positive when it is formed by rotating a line 
counterclockwise and negative when it is formed by clock- 
wise rotation. This is simply another illustration of motion 
in opposite directions. 

EXERCISES 

1. In this exercise the sign indicates the direction of rotation. 
Construct the following angles with ruler and protractor, start- 
ing with the initial line in the horizontal position : -f 30, -(- 45, 
+ 90, +43, +212, -30, -45, -90, -53, -182, -36. 

2. Find the final position of a line which, starting at OX 
(horizontal), swings successively through the following rota- 
tions : + 72, - 38, + 112, - 213, + 336, - 318, - 20, 
+ 228. 

3. Do you see a short cut in finding the final position of 
the line in Ex. 2 ? 




GENERAL .MATHEMATICS 

210. Business relations. Finally, the idea of positive 
and negative numbers may be further illustrated by the 
gain or loss in a transaction ; by income and expenditure ; 
by a debit and a credit account ; by money deposited and 
money checked out; and by the assets and liabilities of 
a business corporation. Thus, a bankrupt company is one 
which has not been able to prevent the negative side of the 
ledger from running up beyond the limit of the confidence 
of its supporters. 

EXERCISES 

1. The assets of a company are $ 26,460, and its liabilities 
are $39,290. What is its financial condition ? 

2. A newsboy having $25 in the bank deposits $10.25 on 
Monday, checks out $16.43 on Tuesday, checks out $7.12 on 
Wednesday, deposits $5 on Thursday, deposits $7.25 on Friday, 
and checks out $11.29 on Saturday. What is his balance for 
the week ? 

3. If a man's personal property is worth $1100 and his real 
estate $12,460, and if his debts amount to $2765, what is his 
financial standing ? 

4. A boy buys a bicycle for $10.25 and sells it for $6. Does 
he gain or lose and how much ? 

211. Addition of three or more monomials. The following 
exercises will help us to see how the addition of monomials 
may be extended. 

EXERCISES 

1. Add the following monomials : 

(a) 2 + 3 + (- 4) + (5). (c) (- 4) + 2 + 3 + (- 5). 

(b) 3 + 2'+ (- 4) + (- 5). (d) (- 5) + (- 4) + 3 + 2. 

2. In what form has the commutative law of addition been 
stated ? (Art. 36.) Does it seem to hold when some of the 
addends are negative? 



POSITIVE AND NEGATIVE NUMBERS 161 

3. Show by a geometric construction on squared paper that 
the commutative law holds when some of the addends are 
negative. * 

HINT. First take a line segment a units long (Fig. 154) and 
add a segment + b units long (Fig. 154) ; then take the segment 
that is + b units long and add the segment Q 

that is a units long. The results should L^ ~ a I 
be the same. 



4. Does a bookkeeper balance the ac- r~~ IfT^ 
count every time an entry is made or does . -.. 

he keep the debits and credits on separate 
pages and balance the two sums at the end of the month '.' 

The process of adding several positive and negative num- 
bers can be explained in detail by the following problem : 

Add - 10, + 50, - 27, + 18, - 22, - 31, + 12. 

Arrange the numb^s as follows : 10 

+ 50 
-27 
+ 18 
-22 
-31 
+ 12 

There are three positive terms, 50, 18, and 12, whose sum is 80. 
There are four negative terms, 10, 27, 22, and 31, whose 
sum is - 90. Adding + 80 to - 90 gives - 10. 

The process consists of the following three steps : 

1. Add all the positive terms. 

2. Add all the negative terms. 

3. Add these two sums by the following process : (a) De- 
termine how much larger the absolute value of one number is 
than the absolute value of the other, (b) Write this number 
and prefix the sign of the greater addend. 



ltli> GENERAL MATHEMATICS 

212. Algebraic addition. The results of the preceding 
exercises show that positive and negative numbers may 
be added according to the f ollo\f1ng laws : 

1. To add two algebraic numbers having like signs find 
the sum of their absolute 'values and prefix to this sum their 
common sign. 

2. To add two algebraic numbers having unlike signs find 
the difference of their absolute values and prefix to it the sign 
of the number having the greater absolute value. 

EXERCISES 

1. Show that the sum of two numbers with like signs is the 
sum of their absolute values with the common sign prefixed. 
Illustrate with a concrete experience. 

2. Show that the sum of two numbers having unlike signs 
but the same absolute value is zero. Illustrate with some fact 
from actual experience. 

3. Find the following sums, performing all you can orally : 

(a) -5 (d)-7 (g) -f -(j)-17f* 
1 - +j +261* 

H 

(b) +5 (e) -Sa (h) -fai (k) + 62z 2 

-28f ae 8 



<<0 + 7 


(f) 


-ftjw 


(i) 


3.16 x 


(1) 


-2.3a;m 2 


1 +5 




16m 




-5.28 a; 




+ 6.5 xm* 


4- 3- 















Find the following sums 

4- -- 6 5. -f 3 6. +51 7. -242 

+ 10 +10 +23 +726 

8 -7 -18 58 

4 - 4 -7 +24 



POSITIVE AND NEGATIVE NUMBERS 163 



8. + 7.5 
+ 12.5 
- 9.5 
+ 2.5 

9. -- Sx 
+ 4z 
+ 17 a; 



10. + 7x 
-lOx 
-I2x 



12. 


+ t 


16. + 7| 


20. 0.5 x 2 




+ | 


7 


0.23 x* 




i 

2 


-i2i 


0.12 a; 2 






- 2J 


0.07 x 2 


13. 


T| 


17. - 12.18 - 


21. +27 a; 2 




+ 1 


- 11.88 


15 ar* 




_ 3 


+ 13.16 


+ 17x 2 




+ 1 


-14.08- 


-12 a: 2 


14. 


+ 3| 


18. +10.05 


22. +23 am 




2 <s 


+ 4.85 


14 xm 




5J 


- 3.25 


17 xm 




2 I 


- 12.35 


+ 20 asm 


15. 


+ 4f 


19. -3.1s 


23. - 18.25 arm 




6 1 


-5.4s 


+ 17.34 aftw 




-6f 


+ 7.2s 


- 19.64 x 2 m 




4 8f 


-3.1s 


+ 21.17 a; 2 m 



11. +24 a 

6a 
- 7a 

3a 



213. Drill exercises. The following exercises constitute 
a drill in determining the common factor of similar mono- 
mials and applying the law for the addition of similar 
monomials (Art. 40). We need to recall that the sum of 
two or more similar monomials is a number whose coefficient 
is the sum of the coefficients of the addends and whose literal 
factor is the same as the similar addends. The exercises 
are the same as a preceding set (Art. 40) except that in 
this case the step in which the coefficients of the addends 
are added involves the addition of positive and negative 
numbers. 



164 GENERAL MATHEMATICS 

EXERCISES 

In each case (1) point out with respect to what factor the 
following terms are similar ; (2) express as a monomial by add- 
ing like terms : 

1. 3y, -6y, 20y, - 35y. 
Solution. The common factor is y. 

The sum of the coefficients is 3 + (- 6) + 20 + (- 35) = - 18. 
Whence the required sum is 18 y. 

2. 5x,7x,9x,+12x,3x. 

3. 7 b, - 12 b, - 9 b, + 11 b, - 13 b. 

4. 9 ab, - 17 ab, - 11 ab, + 13 06. 

5. 8 mnx*, + 12 wmx 2 , 15 wwia 2 , 13 mnx*. 

6. 5 a 2 *, - 7 o%, + 9 a 2 *, - 5 ^b. 

7. 3 ax, 4 ox, 8 az, 7 or. 

8. - 5/,r/, lpq\ - &spq*, ~ Bpf. 

9. az, - 14 z, -bz, +12 z. 
Solution. The common factor is :. 

The sum of the coefficients is a + (- 14) + (-&) + 12. 
Since a and fe are still undetermined, we can only indicate that 
sum thus : a 14 b + 12. 

Whence the sum written as a monomial is (a b 2) z. 

10. mx, + 5 x, 7x, + ex. 

11. m/, - wf, + 5 y 2 , - 12 y 8 , + c/. 

12. tyab, + 4|aJ, 5^ ab, - 6| aft. 

214. Addition of polynomials. We have had numerous 
examples of the addition of polynomials in dealing with per- 
imeters. In applying the principles involved to polynomials 
having positive and negative terms we need to recall that 
in addition the terms may be arranged or grouped in any 
order. Thus, 

2 + 3 + 4 = 3 + 2 + 4 (Commutative Law) 

5 + (- 3) + 4 = - 3 + (5 + 4) (Associative Law) 



POSITIVE AND NEGATIVE NUMBERS 165 

In adding polynomials it is convenient to group similar 
terms in the same column, much as we do in adding 
denominate numbers in arithmetic. 

EXERCISES 

1. Add the following polynomials and reduce the sum to its 
simplest form : 3 yd. + 1 ft. + 6 in., 5 yd. + 1 ft. + 2 in., and 
12 yd. + 3 in. 

Solution. Writing the similar terms in separate columns we have 

3 yd. 1 ft. 6 in. 

5 yd. 1 ft. 2 in. 

12yd. 3 in. 

20yd. 2fE 11 in. 

Note that the common mathematical factors are not yards, feet, 
inches f but the unit common to all of them, or inches. The problem 
may be written as follows : 

3 x 36 + 1 x 12 + 6 

r> x 36 + 1 x 12 + 2 
12 x 36 _ + 3 

20 x 36 + 2 x 12 + 11 

2. Add 9 tj + 3 x + 2 /, o // + 2 x + (> I, and 3 y + 2 x + 8 i. 

9 y + 3 x + 2 i 
5 y + 2 x + 6 ( 
3 y + 2 x + 81 



17 y + Tar 4- 16 i 

3. Add 27 x s 13xy+ 1C if', - 14 a- 8 + 25 xy + 4 ? /, and 



Solution. Writing similar terms in separate columns and adding, 

we have 

27 X s - 13 xy + 16 y 2 

- 14 a- 8 + 25 xy + 4 ?/ 2 



- 6 xz/ + 8 y 2 



166 GENERAL MATHEMATICS 

In the following exercises add the polynomials : 



+ 2b + "c 7. ~6x + 15y-l6z- Sw 

-9/; -2c -Sx + 22y + l6z-12w 

+ 3b 5c 3x+ -ly 5z+ 6w 

2b 3c 9x- 8+ 7z 6w 



5. 2x + 3y + 4 8. 6r-2s + 3t 

5x 6y7z 5r 3t 

4z + 5y + 8s 2r+6s-5t 

-2x-5y-2z 3s + t 



6. 2a + 5b+7c \9\ 2x 3y + z 

3a Sb 5c x+ y 3z 



10. 2x 6b + 13c, llaj + 19ft 30 c, and 5x +4c. 

11. 12A;-10Z + 9m, -27c + 2^-4m, and - 

12. 14 e 3 y + 1 z, 5 e + 5 y 3 z, and 6e + 4:y + 2z. 

13. 24z- 



NOTE. Here certain terms are inclosed in grouping symbols ( ) 
called parenthesis. These indicate that the terms within are to be 
treated as one number or one quantity. Other grouping symbols will 
be given when needed (see pp. 175, 177). 

14. (6*- 

15. 3 



215. Degree of a number. The degree of a number is 
indicated by the exponent of the number. Thus, x 2 is of 
the second degree ; a- 3 , of the third degree ; y\ of the fourth 
degree ; etc. The monomial 3 xyh* is of the first degree 
with respect to x, of the second degree with respect to y, 
and of the third degree with respect to r. 



POSITIVE AND NEGATIVE NUMBERS 167 

216. Degree of a monomial. The degree of a monomial 
is determined by the sum of the exponents of the literal 
factors. Another way of saying this is : The number of 
literal factors in a term is called the degree of the term. 
Thus, 3? is of the second degree ; xy 2 , of the third degree ; 
and 4:ry 2 2 2 , of the fifth degree. 

EXERCISES 

Determine the degree of the following monomials : 

1. 2ajy. 3. 3 ft 4 . 5. 2mxy. 7. ^- 9. rV. 

3.2 z 

2. 2ab s . 4. 5x7/V. 6. 8. rs l . 10. m*x*ifz\ 

m 

217. Degree of a polynomial. The degree of a polynomial 
is determined by the degree of the term having the highest 
degree. Thus, x^y* + x+%y + 5 is of the fourth degree, 
and 5o^ a^ + 7isa third-degree expression. 

EXERCISES 

Indicate the degree of the following polynomials : 

1. x* + 2x 3 - x 4 + 2x* + y + 7. 4. x - 2xif+ if. 

2. x + 2xij + if. ' 5. a; 4 + 4. 

3. y?1xy->ry i . 6. x 5 + x 3 + x* + 1. 

218. Arrangement. A polynomial is said to be arranged 
according to the descending powers of x when the term of 
the highest degree in x is placed first, the term of next 
lower degree next, etc., and the term not containing x 
last. Thus, 2 + x-\-x 8 -\-^3^ when arranged according to 
the descending powers of x takes the form z 3 +3z 2 + #+2. 
When arranged in the order 2 + a; + 3 2 + 3 , the polynomial 
is said to be arranged according to the ascending powers of x. 



168 GENERAL MATHEMATICS 

Find the sura of -3a 2 +2a 3 -4a, -a 2 + 7, 5a 3 -4a 
and -2 3 -7-2a 2 . 

Arranging according to descending powers and adding, we have 
2 a 8 - 3 a 2 - 4 a 

a 2 +7 

5 a 8 - 4 a + 3 

- 2 a 8 - 2 a 2 - 7 

5a 8 -6a 2 -8a + 3 

Check. One way to check is to add carefully in reverse order, as 
in arithmetic. 

A second method for checking is shown by the following : 

Let a = 2. Then we have 

2 a 3 3 a 2 4 a = 4 

a 2 + 7 = 3 

5 a 8 - 4 a + 3 = 35 

- 2 a 3 - 2 a 2 - 7 = - 31 

5a 3 -6a 2 -8a + 3= 3 

The example checks, for we obtained 3 by substituting 
2 for a in the sum and also by adding the numbers obtained 
by substituting 2 for a separately in the addends. 

EXERCISES 

In the following list arrange the polynomials in columns 
either according to the ascending or the descending order of 
some one literal factor. Add and check as in the preceding 
problem. 

1. x 2 + if + xy, x 2 -xy + y*, 14 xy + 3x*-2y' 2 . 

2. 26xi/, -5y* + 12x*, - IQxy - ISxy + 16?f. 

3. 5.3 x 2 - 13.6 xy - 2.3 f, - 0.02 f + 5 xy + 3.2 x 2 . 

4. x 8 -Sx 2 -5x-12, 3x* + 3x-5x* + 8. 

5. 8 a 3 - 2 a 2 + 3 a - 6, - 3 8 + 2 a 2 + a + 7. 

6. 3 ?- 2 + 2 r 8 + 3 r - 5, - r* - 2 r + r s + 1, r 2 - 2. 

7. f s 2 - i - r 2 , - s * _ rs + ^ s f -5r*. 



POSITIVE AND NEGATIVE NUMBERS 169 

8. - 21 oty 4- 40 a? + 55 * 2 ?/ - 14 ^, 

+ 2 cx 2 z + 58 x 2 - 23 y 8 *, 



9. 4 (m + TO) - 6 (m 2 + w 2 ) + 7 (m 3 + n 3 ) - 8(m 4 + n 4 ), 

9 (ra 4 + w 4 ) - 3 (m 8 + w 8 ) + 4 O 2 + re 3 ) - 3 (m + ). 

10. 3 (a + i) - 5 (a 2 + &) + 7 (a - i 2 ) - 5 (a 2 + a& + 6 2 ), 

6 (a + l>) - 3 (a - 6 2 ) + 4 (a 2 + V) + 2 (a 2 + ab + i 2 ). 

219. Subtraction. The following exercises will help to 
make clear the process of subtraction. 

EXERCISES 

1. If the thermometer registers + 24 in the morning and 
-f- 29 in the evening, how much warmer is it in the evening 
than in the morning ? How do you find the result ? 

2. A thermometer registers + 10 one hour and + 14 the 
next hour. What is the difference between these readings ? 

3. If the thermometer registers 2 one hour and + 3 
the following hour, how much greater is the second read- 
ing than the first ? The question might be stated : What is 
the difference between 2 and + 3 ? In what other form 
could you state this question? 

4. If you were born in 1903, how old were you in 1915? 
State the rule you use in finding your age. 

5. The year of Christ's birth has been chosen as zero of 
the time scale in Christian countries. Thus, we record a 
historic event as 60 B.C. or A. D. 14. Instead of using B.C. and 
A.D. we may write these numbers with the minus and plus 
signs prefixed. How old was Caesar when he died if he was 
born in 60 and died in -f- 14 ? 

6. A boy was born in 2. How old is he in + 5 ? Apply 
your rule for subtraction in this case. 

7. Subtract 3 from + 5 ; 4 from + 5 ; 5 from + 5 ; 
- 50 from 4- 25. 



170 GENERAL MATHEMATICS 

8. A newsboy has 410. How much must he earn during 
the day so as to have 850 in the evening? State the rule 
which you use to solve problems of this kind. 

9. A newsboy owes three other newsboys a total of 650. 
How much must he earn to pay his debts and have 200 left? 
Apply your rule for solving Ex. 8 to this problem. 

10. John is $25 in debt, Henry has #40 in cash. How much 
better off is Henry than John ? Apply the rule stated for Ex. 8. 

11. What is the difference between 12 and 20? 5 and 45 ? 
and 20 ? - 1 and 20 ? - 5 and 25 ? - 12 and 18 ? . 

12. Interpret each, of the parts of Ex. 11 as a verbal problem. 

13. Through how many degrees must the line OI^ turn 
(Fig. 155) to reach the position OR 2 ? 

14. Archimedes, a great mathema- 
tician, was born about the year 287 
and was slain by a Roman soldier 
in 212 while studying a geometrical 

figure that he had drawn in the sand. FIG. 155 

How old was he ? 

15. Livy, a famous Roman historian, was born in 59 and 
lived to be 76 yr. old. In what year did he die ? 

16. Herodotus, the Greek historian, sometimes called the 
Father of History, was born in 484 and died in 424. At 
what age did he die ? 

220. Subtraction illustrated by the number scale. In 
subtracting 4 from 6 we find what number must be added 

-8 -7 -6 -5 -4 -3 -2 -1 +1 4-2 +3 4-4 -H5 4-6 +7 4-8 
< i I I I I I I I I I I I I i i i i > 

FIG. 156. THE NUMBER SCALE 

to the 4 (the subtrahend) to get the 6 (the minuend). 
On the number scale (Fig. 156) how many spaces (begin- 
ning at 4) must we count until we arrive at 6 ? 




POSITIVE AND NEGATIVE NUMBEKS 171 

ILLUSTRATIVE EXAMPLES 

1. Subtract 2 from 3. 

Solution. Beginning at 2 we need to count 5 spaces to the 
right (positive) to arrive at 3. Hence, subtracting 2 from 3 equals 5. 
Note that we might have obtained the result by adding + 2 to 3. 

2. Subtract + 5 from 2. 

Solution. Beginning at 5 on the number scale we need to count 7 
to the left (negative) to arrive at 2. Hence, subtracting + 5 from 
2 equals 7. Note that we could have obtained the same result 
by adding 5 to 2. 

This exercise may be stated as a temperature problem ; namely, 
What is the difference between 5 above zero and 2 below zero ? 

3. Subtract 8 from 2. Interpret as a verbal problem. 

Solution. Beginning at 8 on the number scale we need to count 
6 to the right (positive). Hence, subtracting 8 from 2 equals + 6. 
Notice that the same result is obtained if + 8 is added to 2. 

These examples show that since subtraction is the reverse 
of addition, we can subtract a number by adding its opposite. 
Thus, adding $100 to the unnecessary expenses of a firm 
is precisely the same as subtracting $100 gain, or, on 
the other hand, eliminating (subtracting) $1000 of lost 
motion in an industrial enterprise is adding $1000 to the 
net gain. 

It is convenient for us to make use of this relation, for 
by its use there will be no new rules to learn, but merely 
an automatic change of sign when we come to a subtraction 
problem, and a continuation of the process of addition. 

221. Algebraic subtraction. The preceding discussion 
shows that subtraction of algebraic numbers may be changed 
into algebraic addition by the following law: To subtract 
one number from another change the sign of the subtrahend 
and add the result to the minuend. 



172 GENERAL MATHEMATICS 

Thus, the subtraction example 

+ 7a 

- 3a 

+ 10 a 

may be changed to the addition example 

+ la 
+ 3a 
+ 10 a 

EXERCISES 

Subtract the lower number from the upper number in the 
following. Illustrate Exs. 1-11 with verbal problems. 



1. 


29 


9. -14 


17. +x 


25. 


+ 0.81 cc 2 




-10 


- 3 


x 


f 


-2.62x 2 


2. 


-55 


10. +14 


18. x 


26. 


-.660 




-15 


+ 3 


+ # 




-.840 


3. 


-65 


11- -T 9 o 


19. +x 


27. 


+ 0.82 r 




+ 15 


. 2 


X 




+ 2.41 r 


4. 


+ 18 


12. -5 a- 2 


20. +5.74 a; 8 


28. 


- 3.34 a 




+ 24 


+ a-* 


-6.26 a- 8 




+ 5.37 a 


5. 


A 


13. -126 8 


21 -3.15 a 2 


29. 


+ 2.04y 




T2 


7b 3 


.34 a 2 




-4.23y 


6. 


3 
"ff 


14. - 9r 


22. +6(a + &) 


30. 


+ 8.92 a; 




~ T 


- 14 / 


8(a + J) 




+ 9.17 -jr 


7. 


3 


15. + 7 2 


23. _5(a-6) 


31. 


- 7.42 z 




14 


-14 a 2 


3 (a ft) 




- 3.71 


8. 


14 


16. -7m 8 


24. -4.36* 


32. 


-2.417/ 




- 3 


+ 3m 8 


8.64* 




+ 8.62?/ 



POSITIVE AND NEGATIVE NUMBERS 173 

TRANSLATION INTO VERBAL PROBLEMS 

33. Translate each of the following subtraction exercises 
into a verbal problem, using the suggestion given : 

+ 8246 

(a) As assets and liabilities : 

+ 5 

(b) As gain or loss : 



I 

(c) As debit or credit : 



48 
(d) As an angle problem : 



-27 

14 



+ 22 
(e) As an age problem (time) : 

(f ) As line segments on the number scale : 

246 

(g) As a bank account : 



40 

(h) As a latitude problem : 

-(- Zo 

i go 
(i) As a longitude problem: 

~l~ 75 

i 2 
(j) As a problem involving forces : 

222. Subtraction of polynomials. When the subtrahend 
consists of more than one term the subtraction may be 
performed by subtracting each term of the subtrahend 
from the corresponding term of the minuend. 

For example, when we wish to subtract 5 dollars, 3 quarters, and 
18 dimes from 12 dollars, 7 quarters, and 31 dimes, we subtract 5 
dollars from 12 dollars, leaving 7 dollars ; 3 quarters from 7 quarters, 
leaving 4 quarters ; and 18 dimes from 31 dimes, leaving 23 dimes. 



174 GENERAL MATHEMATICS 

The subtraction of algebraic polynomials is, then, not 

different from the subtraction of monomials and may 

therefore be reduced to addition, as in the following 
two examples, which are exactly equivalent: 

SUBTRACTION ADDITION 

7 2 - 14 ab - 11 P 7 a 2 - 14 ab - 11 tf 

+ 5 a 2 + 3 ab + 3 ft 2 - 5 a 2 - Sab- 3 b* 

2 a* - 17 ab - 14 ft 2 2 a 2 - 17 // - 14 ft 2 

The student should change the signs of the subtrahend 
in the written form until there is no doubt whatever as 
to his ability to change them mentally. The example will 

appear as follows: 

2 + 2 ab + // 

- + 

+ a 2 - 2 ab + fe 2 

4o6 

NOTE. The lower signs are the actual signs of the subtrahend. 
They are neglected in the adding process. 

Numerous verbal problems have been given with the 
hope of giving a reasonable basis for the law of subtraction. 
The student should now apply the law automatically in the 
following exercises. 

EXERCISES 

Subtract the lower from the upper polynomial : 

1. 4a?-3ab + 6b 2 3. x 3 + 3x 2 y + Sxy 2 + y 3 

4:a?-5ab-4:b 2 - 7 x z y + 3 xy* + y* 



2. x 2 -5zy+ if 4. 2mn 2 + 5m s n+ 6 

-3x 2 -4a;//-3y 2 - 7 mri* - 4 m*n + 18 

5. From 10 xy 5 xz -j- 6yz subtract 4 xy 3 xz + 3 yz. 

6. From 16 x - 5 mx* + 4 m* subtract 7x* 4 mx* + 12 m 3 . 

7. From 2a?-2a?b + al> 2 -2b 3 subtract a 3 - 3 a?b + aft 2 - 6 s . 



Subtract as indicated, doing as much of the work as possible 
mentally. 

8. (4r 8 6 r 8 * + 10 s 8 - 6 rs 8 ) - (2 r 8 + 6 r 2 * + 4 s 8 -f- 3 rs 2 ). 

9. (_ 8 m *pq - 4 m s p -10mV) -(-6m*p-8m?pq - 15 mV)- 

10. (15 x s - 12 afy + Ty 8 ) - (- 11 z 8 + 8 a"y - 5 y 8 ). 

11. ( a 8 - 3J aft 8 - 3 a 8 *) - (- f a& 2 + 5^ 2 6 - 3 a 8 ). 

12 . (5f rst - 7 J r 8 * - 8f s 8 * - 6 J * 8 /-) - (4 J s 2 * + 3| ** + 7 J A). 

13. (2.3 aW - 4.6 a 4 6 8 + 8.7a 6 * 2 ) -(-1.1 aW - 2.1 a 6 * 8 - 3a 3 6 4 ). 

14 . (3 x 2 - 4 x + 5) -f (2 x 2 + 5 x - 3) - (- 2 a 2 + 3 x + 6). 

15. (5.2 ofy - 41xy + 2 if) - (31 afy + 3.2 xy + 5 y 2 ). 

16. (2.42 a 2 6 2 + 5 ab + 6) - (3.12 a 8 ^ 2 ai - 9). 

17. (3 a6 3 - 3 afe 8 ) - (-2 a& 8 + 3 a 3 - 4 a5c 8 ) - (- 4 a 8 - 

18. (5x 2 +2a-// + 3y 2 ) + (2* 2 -5a;y-/)-(9x 2 

19. Compare the signs. of the terms of the subtrahend in the 
foregoing exercises before and after the parenthesis are removed. 

20. State a rule as to the effect of the minus sign preceding 
a polynomial in parenthesis. 

21. What is the rule when the plus sign precedes a poly- 
nomial in parenthesis ? 

223. Symbols of aggregation. It has been found very 
convenient to use the parenthesis for grouping numbers. 
Such a symbol indicates definitely where a polynomial 
begins and ends. Other symbols used with exactly the 
same meaning and purpose are [ ] (brackets) ; { } (braces) ; 
and " " (vinculum). Thus, to indicate that a + b is 
to be subtracted from x + y we may use any one of 
the following ways : (# + #) (# + ft), [x + y\ [a + ft]. 
{x-}- y} {a + ft}, or x+y a + b. The vinculum is like 
the familiar line separating numerator and denominator 

,. . . .2 a + ft 

oi a fraction, as in - or - 

o a ft 



170' (JEMERAL MATHEMATICS 

Sometimes the symbols are inclosed one pair within 
another; thus, 19 - (16 - (9 - 2)}. 

In an example like the preceding the common agree- 
ment is to remove first the innermost parenthesis. First, 
2 is to be subtracted from 9, then the result, 7, is to be 
subtracted from 16. This result, 9, is in turn to be sub- 
tracted from 19 ; whence the final result is 10. 

EXERCISES 

1. (live the meaning of the following: 
(a) 15 -{4 + (6 -8)}. 



( C ) _ 5 x _ [_ 7 x _ {2x- 
(d) 3 (a; + y) - 5{x - 2x-3y}. 

2. Keep definitely in mind the rules governing the effect of 
a minus or a plus sign before a grouping symbol. Perform the 
following indicated operations and simplify the results : 

(a) 12-{5-(-2x-5)}. 

(b) 17 - {- 12 x - 3 x - 4}. 

(c) 4 a 2 - (a 2 - 3 a 8 + 3 a 2 - a 8 }. 

(d) 2e-[6e-36-4e-(2e-46)]. 

(f ) 15 a? - {- 3 x 2 - (3 x 2 + 5)} - (20 a 2 + 5). 

SUMMARY 

224. This chapter has taught the meaning of the follow- 
ing words and phrases : positive number, negative number, 
algebraic numbers, absolute value of a number (or numerical 
value), degree -of a number, degree of a monomial, degree 
of a polynomial, descending power, ascending power. 



POSITIVE AND NEGATIVE NUMBEKS 177 

225. The following symbols were used : -f (plus sign) 
and (minus sign) for positive and negative number re- 
spectively, ( ) (parenthesis), [ ] (brackets), { } (braces), 
and ~~ ~ (vinculum). 

226. Positive and negative numbers have been illustrated 
by game scores, directed line segments, latitude, longitude, 
time, the number scale, forces, the thermometer, angles, 
profit and loss, debit and credit, assets and liabilities, 
deposits and checks. 

227. The sum of two algebraic numbers with like signs 
is the sum of their absolute values with their common 
sign prefixed. 

The sum of two algebraic numbers with unlike signs 
equals the difference of their absolute values with the sign 
of the number having the greater absolute value prefixed. 

The sum of three or more monomials is found most 
easily by the following method : (1) add all positive terms, 
(2) add all negative terms, (3) add the two sums obtained. 

228. To add polynomials add the similar terms (write 
similar terms in the same column). 

229. To find the difference of two numbers change the 
sign of the subtrahend and add. 

230. A parenthesis is used for grouping. If preceded 
by a plus sign, it may be removed without making any other 
changes ; if preceded by a minus sign, it may be removed 
if the sign of every term within the parenthesis is changed. 



CHAPTER IX 



+ 4 



POSITIVE AND NEGATIVE NUMBERS. MULTIPLICATION 
AND DIVISION. FACTORING 

231. Multiplication. The laws of multiplication of num- 
bers having plus or minus signs are easily applied to a 
considerable number of interesting problems. These laws 
are illustrated in the following examples: 

ILLUSTRATIVE EXAMPLES 

1. Find the product of (+ 4) and (+ 2). 

Solution. Geometrically we interpret this as follows : Take a 
segment + 4 units long and lay it off two times to the right of 
zero on the number scale ; that 
is, in its ewn direction (Fig. 157). 
Thus, (+2) (+4) = + 8. 

2. Find the product of 
(-4) and (+2). 

Solution, Geometrically this 
means : Take a segment 4 units 
long and lay it off two times to 
the left of zero on the number 
scale ; that is, in its own direction 
(Fig.158). Thus,( + 2)(-4) = -8. 

3. Find the product of 
(+ 4) and (- 2). 

Solution. Geometrically we in- 
terpret this as follows: Take a 
segment + 4 units long and lay it off two times to the left of zero ; 
that is, opposite its own direction (Fig. 159). Thus, ( 2) (+ 4) = 8. 



i i i i i i i i i i i i i i i i 
+ 8 r ^ X 



-4 



FIG. 157 



4 



L -4 



O 
FIG. 158 

-4 J 



I I I I i I 



O 
FIG. 159 



POSITIVE AND NEGATIVE NUMBEKS 179 

4. Find the product of (- 4) and (- 2). 

Solution. If the first factor were a positive 2, then we should 
interpret this geometrically by laying off 4 twice, obtaining a 
line segment 8 units long (see 
OR V in Fig. 160) just as we did in 
Ex. 2. But since it is a negative 2, 
we lay it off not in the direction of 
OR l but in the opposite direction; p , 

namely, OR (see Fig. 161). Thus, 



Note that in 
this last case, as 

in Ex.1, the signs 

FIG. 161 

of the multipli- 
cand and the multiplier are alike, and the product is positive ; 
while in Exs. 2 and 3 the signs of the multiplicand and 
multiplier are unlike, and the product is negative. 

EXERCISES 

1. Find geometrically the products of (+ 2) (+ 5); (- 2)(+5); 
(+2) (-5); (-2) (-5). 

2. State the law of signs for the product of two algebraic 
numbers as suggested by the preceding work. 

232. The law of signs for multiplication. The law of 

signs for multiplication is as follows: 

The product of two factors having like signs is positive. 
The product of two factors having unlike signs is negative. 

EXERCISES 

Find the value of the following products, using the law of 
signs. Illustrate the first ten geometrically. 

1. (+3) (+5). 3. (-3) (+5). 5. (-2) (+3). 

2. (- 3)(- 5). 4. (+ 3)(- 5). 6. (- 2)(- 3). 



180 



GENERAL MATHEMATICS 



7. (-2) (+7). 

8. (+2) (+7). 

9. (9) (-3). 
10. (-4) (-a). 



11. (2)(-). 

12. (-!)(-!). 

13. (-2*) (-3). 

14. 2a/>-3. 



15. (-3) (-5 a). 

16. (-f)( 

17. (-f)( 

18. -2o- 




FIG. 162. THE LAW or MULTI- 
PLICATION ILLUSTRATED BY THE 
BALANCED BEAM 



233 ! . Law of multiplication illustrated by the balance. 
The law of signs may be illustrated with a balanced bar 
(Fig. 162). A light bar is balanced at M. The points r v r 2 , 
etc. represent pegs or small 
nails driven at equal distances. 
We shall speak of r v r v etc. 
as " first right peg," " second 
right peg," etc. and of l v 1 2 , 
etc. as " first left peg," " sec- 
ond left peg," etc. with the bar 
in a position facing the class as 
in Fig. 162. The weights, w, 
are all equal ; hence we shall 
merely speak of them as " two weights," " three weights," 
etc. instead of mentioning the number of ounces or grams 
contained. In Experiments 1-3 the string over the pulley 
is fastened on the first left peg. 

EXPERIMENTS 

1. Hang two weights on / r This tends to turn the bar. How 
many must be attached to the hook H to keep the bar level ? Hang 
three weights on / r What do you notice about the turning tendency 
as compared with the first case ? Answer the same question for four 
weights on 1 3 . 

1 The entire article may be omitted at the teacher's discretion. The 
device has, however, proved useful in the hands of many teachers. The 
apparatus may be bought at several of the large book companies or, 
better still, made in the shop by a member of the class, using a part of 
a yardstick for the lever and small nails for pegs. 



POSITIVE AND NEGATIVE NUMBERS 181 

2. Hang one weight on l r How many must be placed on the hook 
to keep the bar level? Hang one weight on / 2 ; remove it and hang 
one weight on / 3 ; on / 4 ; and so on. What do you notice about the 
turning tendency in each case ? What two things does the turning 
tendency seem to depend on ? 

3. With the string passing over the pulley fastened to /j ho.w 
many weights must be put on the hook to balance two weights 
on / 8 ? three weights on 1 2 ? one weight on l t ? two weights on Z 4 ? 
three weights on / 4 ? 

4. Repeat Experiments 1~3 for the pegs on the right side, with the 
pulley string fastened to r r What seems to be the only difference? 

Results of experiments. The experiments show that 

1. The turning tendency (Jorce) varies as the number of 
weights hung on a peg on the bar. Thus, the more weights 
hung on any peg, the stronger the force. 

2. The turning tendency also varies as the distance of 
the peg from the turning point. 

3. The turning tendency is equal to the product of the 
iveights multiplied by the distance of the peg on which the 
weight hangs from the turning point. 

4. When a weight is hung on a right peg, the bar turns in 
the same direction as the hands of a clock ; when a weight is 
hung on a left peg, the bar rotates in a direction opposite to 
the hands of a clock. 

234. Signs of turning tendency ; weight ; lever arm. It 
is conventionally agreed that when the bar turns counter- 
clockwise (as you face it), the turning tendency is positive ; 
while if the bar rotates clockwise, the turning tendency is 
negative. 

Weights attached to the pegs are downward-pulling 
weights and are designated by the minus sign. Weights 
attached at H pull upward on the bar and are designated 
by the plus sign. 



182 GENERAL MATHEMATICS 

The distance from the turning point to the peg where 
the weight, or force, acts will be called the lever arm, or 
arm of the force. Lever arms measured from the turning 
point toward the right will be marked + ; those toward 
the left, . For example, if the distance from M to peg i\ 
is represented by + 1, then the distance from M to r 3 will 
be + 3 ; the distance from M to / 2 will be 2 ; and so on. 

235. Multiplication of positive and negative numbers. By 
means of the apparatus (Fig. 162) the product of positive 
and negative numbers is now to be found. 

ILLUSTRATIVE EXAMPLES 

1. Find the product of (+ 2) (- 4). 

Solution. AVe may interpret this exercise as meaning, Hang four 
downward-pulling, or negative, weights on the second peg to the 
right (positive). The bar turns clockwise. The force is negative; 
hence the product of (2) ( 4) is 8. 

2. Find the product of (- 2) (- 4). 

Solution. Hang two downward-pulling, or negative, weights on the 
fourth peg to the left (negative). The bar turns counterclockwise. 
The force is positive ; hence the product of ( 2) ( 4) is + s. 

3. Show that (+ 3) (+ 4) = + 12. 

HINT. Fasten the string over the pulley to the fourth peg to the 
right and hang three weights on the hook. 

4. Show 'that (-3) (+2) = -6; that (+ 2) (- 3) = - 6. How 
does the beam illustrate the law of order in multiplication ? 

5. Compare the results of Exs. 1-4 with the law of signs in 
multiplication (Art. 232). 

It is hoped that the law of signs is made reasonablv clear 
by means of these illustrations. The student should now 
proceed to ajjply the law automatically. 



POSITIVE AND NEGATIVE NUMBERS 183 



EXERCISES 



State the products of the following, doing mentally as much 
of the work as possible : 



1. (+4) (-6). 

2. (-4) (+6). 

3. (+4) (+6). 

4. (-4) (-6). 

5. (+2) (+5). 

6. (+3) (-4). 

7. (-5) (-2). 

8. (-3) (-7). 

9. (-5) (+6). 



11. (-3.1) (-5). 21. 

12- (-f)(f). 22. (-6X-S). 

23. (-8)(- 

24. (-c)(- 

25. (- 

26. (- 

17. (+6j-)(+6j). 27. (51) (-^ 2 ) 

18". (+6j)(-6-i). 28. (-9)(+x 2 



16. _6- 



19. (_6i)(+6-i-). 



10. (-12) (-13). 20. 



29. (-1) 3 . 

30. -2 3 . 



236. Multiplication by zero. The product of 3 x means 
+ + = 0. 

EXERCISES 

1. Show geometrically that ax 0=0. 

2. Show by the beam (Fig. 162) that a x 0=0; that x a=0. 

3. State a verbal problem in which one of the factors is 
zero. In general both a x and x a equal zero. Hence the 
value of the product is zero when one of the factors is zero. 

4. What is the area of the rec- 
tangle in Fig. 163 ? How would the area 
change if you were to make the base 
smaller and smaller ? What connection 
has this with the principle a x = ? 

5. How would the area of the rec- 

tangle in Fig. 163 change if I> were not changed but a were 
made smaller and smaller ? What does this illustrate ? 



b 
FIG. 163 



184 GENERAL MATHEMATICS 

237. Product of several factors. The product of several 
factors is obtained by multiplying the first factor by the 
second, the result by the third, and so on. By the law of 
order in multiplication the factors may first be rearranged 
if this makes the exercise easier. This is often the case in 
a problem which involves fractions. 

EXERCISES 

1. Find the value of the following products : 
(a) (+2) (-3) (-5) (-4). 



00 (- -BX-f) (-!!)() 

2. Find the value of (-1) 2 ; (- 1) 3 ; (-1) 4 ; (-2) 2 (-2) 3 
(-2)*; (-2) 5 ; (- 3) 2 (- 3) 8 (- 3)*; (_4) 2 (-4) 8 . 

3 . Find the value of 3 a- 4 5 x s + x* 12 x 5 when x 2. 

4. Find the value of x 8 3x*y + Secy 2 + y* when x = 3 
and y = 2. 

5. Find the value of z 8 + 3 x 2 + 3 x + 1 when x = 10. 

6. Compare (- 2) 3 and - 2 s ; - 3 s and (- 3) 8 ; (- 2) 4 and 
- 2 4 ; (- a) 8 and - 8 ; (- a) 4 and - a*. 

7. What is the sign of the product of five factors of which 
three are negative and two are positive? of six factors of 
which three are negative and three are positive ? 

8. What powers of 1 are positive ? of 2 ? of x ? 
State the rule. 

238. Multiplication of monomials. Find the product of 



The sign of the product is determined as in Art. 232 and is found 
to be + . 

By the law of order in multiplication the factors may be arranged 

as follows : , QN 

2(- 3) (- o) zxxxyyy, 

which is equal to 3 



POSITIVE AND NEGATIVE NUMBERS 185 

Hence, to find the product of two or more monomials 

1. Determine the sign of the product. 

2. Find the product of the absolute values of the arith- 
metical factors. 

3. Find the product of the literal factors. 

4. Indicate the product of the two products just found. 

EXERCISES 

Simplify the following indicated products, doing mentally as 
much of the work as possible : 



2. (-2*) (-3*) (4*). 

3. (5 xV 2 ) (- 3 *V) ( 2 ay* 3 ) (- 7 ay* 2 ). 

4. (- 2 a 2 bcd) (5 aPcd) (- 7 abc 2 d) (- 2 alcd*). 

5. ( 3 mnx) ( 5|- m*nx) ( 2 oj). 



7. (- 3J 06) (- 5 J a 2 ^ 2 ) (+ 8 oft). 

8- (5i?V)(l^X 

9. (I Vty (6 jac^^) ( 

10. (3 ^) (6 m 2 m ^ } ( 

11. (1.3 xV*)(- 2 aV u * u 

12. (1.1 a-y 2 ) (1.1 mxy 2 ) (10 i 2 o; 2 ). 

13. ( _ 2 yX-3) 2 (-4)(2) 2 . 

14. (- 3) 2 (- 2) 2 (- I) 3 . 

15. (5)(-4)(3) 2 (0)(2) 2 . 

16. (- a)\- a}\- a 2 ) (- a 3 ). 

17. (aj + y)(a5 + y)(a5 + y) 6 . 

18. 3 (aj + y) 8 (x - 7/) 2 (x + y)'(x - y}*. 

19. Find the value of 1+ (- 5) (- 3) - (- 2) (4) - (- 3) (3; 



186 GENERAL MATHEMATICS 

20. What is a short method of determining the. sign of the 
product containing a large number of factors ? 

NOTE. It is agreed that when an arithmetical expression contains 
plus or minus signs in connection with multiplication or division 
signs, the multiplication and division shall be performed first. This 
amounts to the same thing as finding the value of each term and then 
(tili/in</ in- s it lit racling as indicated. 

239. Multiplication of a polynomial by a monomial. We 
shall now see how the process of algebraic multiplication 
is extended. 

INTRODUCTORY EXERCISES 

1. Keview the process of finding the product of a(x + y + z) 
in Art. 122. 

2. Illustrate by a geometric drawing the meaning of the 
product obtained in Ex. 1. 

3. How many parts does the whole figure contain ? 

4. What is the area of each part ? 

The preceding exercises serve to recall the law that a 
polynomial may be multiplied by a monomial by multiplying 
every term of the polynomial by the monomial and adding the 
resulting products. 

DRILL EXERCISES 

Find the products as indicated and check by substituting 
arithmetical values for the literal numbers : 



Solution. a 2 - 2 ab + 3 b* = W 

_ 3_a= _6 
3 a 3 - 6 a 2 6 -f 9 a& 2 = 114 



Check. Let a = 2 and b = 3. Then the same result is obtained by 
substituting in the product as by substituting in the factors and 
then multiplying the numbers. Note that the check is not reliable 



AND NEGATIVE NUMBERS 1ST 

if we let a literal number in a product containing a power of that 
literal number (as a: in a product a; 5 4 x) equal 1, for if x = 1, then 
x 5 4 x also equals x 3 4 x, x 2 4 x, x 9 4 x, etc. Explain. 

2. 5x(2x* -3x -7). 

3. 

4. 

5. 5.1 4 (i 

6. 

7. (?V - 3 mV + 4 mV - 9 wV) 3.5 ?V. 

8. 

9. 
10. 
11. 



240. Product of two polynomials. In Art. 126 we found 
the product of two polynomials to be the sum of all the partial 
products obtained by multiplying every term of one polynomial 
by each term of the other. After reviewing briefly the case 
for positive terms we shall proceed to interpret the above 
law geometrically even when negative terms are involved. 

ILLUSTRATIVE EXAMPLES 

1. Find the product of (c + d)(a + It). 

Solution. The area of the whole rectangle in Fig. 164 is expressed 
by (a + &) (c + d). The dotted line suggests a M d 

method for expressing the area as the sum of 
two rectangles ; namely, a (c + c?) + b (c + d). 
If we use the line MN, the area may be 
expressed as the sum of four rectangles ; 
namely, ac + ad + be + bd. Each expression c wo 

equals the area of one of the rectangles; hence FIG. 164 

(a + 6) (c + d) = a(c + d) + b(c + d) = ac + ad + be + bd. 



1.88 



GENERAL MATHEMATICS 



3x y 

FIG. 165 




G a B 



FIG. 166 



2. Illustrate, by means of Fig. 165, the law for the 
multiplication of two polynomials. % x y 

3 . Find the product of (a t>)(c + d). 

Solution. In this case one of the factors 
involves a negative term. 

The product (a b)(c + d) is repre- 
sented by a rectangle having the dimen- 
sions (a - b) and (c + rf) (Fig. 166). The 
rectangle ABEF=ac + ad. Subtracting 
from this the rectangles be and bd, we 
obtain the rectangle A BCD. 

Therefore (a b) (c + d) = ac + ad 
be bd, each side of the equation repre- 
senting the area of rectangle A BCD. 

* 4 . Findthe product of (a b) (a fy . 

Solution. Let ABCD (Fig. 167), repre- 
sent a square whose side is (a 6) feet. 
Show that the area of ABCD equals 
EFGC + GHIB - FKDE - KHIA. 
Then 

(a - ?/) (a - b) = (a - b) 2 Why ? 

= a 2 + 6 2 ab ab 
= a 2 - 2 ab + IP. 

*5. Sketch a rectangle whose area is (m + ri) (r s) ; whose 
area is 24 b 2 6 be. 

DRILL EXERCISES 

Apply the law of multiplication to two polynomials in the 
following exercises. Check only the first five. 



Solution. 



D 








(a-b)' 




A 




B 






b 1 


f 


: G 


1 




FIG. 167 





x z + 2 xy + y- 
x + y _ 
x 3 + 2 x' 2 y + xy 2 



+ 2 xy* + y* 



x* + 3 x*y + 3 xy* + y* 
Check by letting x = 2 and y = 3. 



POSITIVE AND NEGATIVE NUMBERS 189 

2. (rs + tm) (rs - tm). 7. (2 a + 3 ft) (2 a - 3 4). 

3. (a 8 + ax 3 + a s ) (a + a). 8. (i aft - ftc) (f aft + f ftc). 

4. (a 2 + 4 a; + 3) (a -2). 9. (a + 6 - c) 2 . 

5. (cc 2 -3cc + 5)(2x + 3). 10. (a - b + c - df. 

6. (k 2 + 3 Ar + 1) (A - 2). 11. . (- 2 a + 3 1> - 4 c) 2 . 

12. Comment on the interesting form of the results in 
Exs. 9-11. 

13. (0.3 a + 0.4 b - 0.5 c) (10 a - 30 b + 40 c). 

14. (2 if - 12 zy + 5 x 2 ) (2 if - 5 .r 2 ). 

15. o 2 + *z/ + z/ 2 ) (a - y) 0* + y;. 

16. (9x 2 

17. (x + 

18. Comment on the form of the results in Ex. 17. 

19. (r 2 + rs - s 2 ) (r 2 + rs + .<?). 

20. (Sr 2 + 5 r + 6) (3 r 2 + 3 .* - 6). 

21. (3x + 2.y) 3 -(3x-2y) 8 . 

22. (3 x 2 + T/ 2 ) 2 - (3 x z - iff -x*(x- 

23. (2a-3^) 2 -(2a + 36) 2 + (2 

24. (0.3 a - 0.4 4) 3 - (0.5 a + 0.6 i) 2 

- (0.3 a - 0.4 ft) (0.3 a + 0.4 ft). 



26. (5 + 3) 2 - (5 - 3) 2 - (5 + 3) (5 - 3). 

27. Why may 3527 be written 3 10 s + 5 10 a + 2 . 1 + 7 ? 

28. Multiply 352 by 243. 

HINT. Write in the form 3 10 3 + 5 10 + L> 
2 10 2 + 4 10 + 3 

29. Write 56,872 as a polynomial arranged according to the 
descending powers of 10. 

30. Find the product of 5 and 3427 by the method suggested 
in Ex. 28. 



190 GENERAL MATHEMATICS , 

241. Product of two binomials. We shall now see how 
the algebraic product of two binomials may be obtained 
automatically. The following exercises will help the stu- 
dent to discover and understand the method. 

EXERCISES 
Find by actual multiplication the following products : 

1. (2x + 3)(4z + 5). 4. (4x + 6) (4 a; + 5). 
Solution. 2* + 3 5. (3* -2) (3* -2). 

4x+5 6. (x + 2) (* + 9). 

8* 2 + 12* 7. (2a;+l)(aJ4-6). 

_ +l* + lr ' 8. (b + 3) (6 + 5). 

8 x 2 + 22 x + 15 

9. (a -7) (a -3). 

2. (3 + 5)(2a-8). 10 ' (3 * + 8) (a + 2). 

11. (3x + 4)(2a;-3). 
Solution. 3 a + 5 

2a -8 12 ' (*-3)(*-10). 

13. (a- 8 - 9) (*" + 9). 



- 24 a - 40 14. (or 2 - 5) (aj 2 + 10). 

6a 2_14 a _ 40 15> (3 a; -5) (4 a; -2). 

16. (2y 3)(5y 8). 
3. (o ?/ + 4) (o ?/ 4). 

17. (i 
Solution, o M + 4 1 o / 



-My -16 20. (3a; + 4y)(3a;-'4y). 

-16 21. (4a + 26)(7a-5i). 

22. 53 x 57. 
Solution. 53 x 57 = (50 + 3) (50 + 7) 

= 50 3 + (7 + 3)50 + 21. 

23. 61 x 69. 24. 52 x 56. ' 25. 37 x 33. 

26. Can you see any way of formulating a rule for finding 
the products of two binomials ? 



POSITIVE AND NEGATIVE NUMBERS 191 

If we agree to use the binomials ax 4- b and ex 4- d to 
represent any two binomials where a, b, e, and d are 
known numbers like those in the products above, then we 
may discover a short cut in multiplying ax 4- b by ex 4- d. 

ILLUSTRATIVE EXAMPLE 

Find the product of (ax 4- V) and (<-x -\- <Z). 
Solution. ax + b 

Y 

CX + d 



bcx 
. + adx + bd 



acx 2 + (be + ad ) x + bd 

The arrows show the cross-multiplications or cross-products whose 
sum is equal to the middle term. It is seen that the first term of the 
product is the product of the first terms of the binomials, that the last term 
is the product of the last terms of the binomials, and that the middle term 
is the sum of the tivo cross-products. 

EXERCISES 

Using the rule stated above, give the products of the fol- 
lowing binomials : 

1. (2 a + 3) (3 a 4- 5). 6. (3 a - 2fi)(3a - 2 ft). 
Solution. The product of the 7. f x _ 7) (4 x + 9). 

first terms of the two binomials 

is 6 a 2 , the product of the last 

terms is 15, and the sum of the 9. (x + *>)(x + 8). 

cross-products is 19 a. Therefore , ^ / 7 9 v. 

the product is 6 a + 19 a + 15. 

2. (4,, + 3) (2. + 1). 11. (4, + 3)(3,-4). 

3. (2s -7) (3* + 2). 12 - (*4-9&)(aj + 7ft). 

4. (3 x 4- 4) (3x4- 4). 13. (2 aj + 4 y) (3 x - 5 y). 

5. (7 x - 2) (7 x + 2). 14. (5 a + 4) (4 a - 2). 



192 GENERAL MATHEMATICS 

15. (7a + 26)(7o-2i). 18. (3 a - 7 ft) (3 a - 7 6). 

16. (5a + 4i)(5 + 46). 19. (6 xy + 2) (3 xy - 5). 

17. (3 a: + 2) (12 a: -20). 20. (7 ab + 5c) (60* - 8c). 

21. Do you notice anything especially significant about the 
product of two binomials that are exactly alike? Explain 
by using the product of x + y and a- + y, a b and a b 
(compare with Ex. 1, Art. 127). 

22. Do you notice anything especially significant about the 
product of two binomials that are the same except for the 
signs between the two terms ? Explain by using the product 
of ra + n and m n. 

23. Try to formulate a rule for obtaining automatically the 
products referred to in Exs. 21 and 22. 

242. Special products. We have seen in Art. 241 how 
the multiplication of two binomials may be performed 
automatically. Such products are called special products. 
The student should observe tha f , Exs. 21 and 22, Art. 241, 
furnish examples of such products. For example, the 
product of x'+y and x -+- y is equal to x 2 + 2 xy + y*, and 
is called the square of the sum of x and y\ while the 
product of a b and a b is equal to a 2 2 db + 6 2 , and 
is called the square of the difference of a and b. Further, 
the product of m + n and m n is equal to m z n*, and 
is called the product of the sum and difference of m and n. 

EXERCISES 

1. Find automatically the following special products and 
classify each: 

(a) (x + 3)(x + 3). (c) (2x + 4) (2 a; + 4). 

(b) (y_2)(y-2). (d) (3* -6) (3* -6). 

(e) (2x + 4) 2 . (g) (2 x + 4 y) 2 . (i) (2 a + 4 ft). 

(f ) (4 x - 2) 2 . (h) (5 x - 2 y}\ (j) (3 a - 2 ) 2 . 



193 

2. State a rule for finding automatically the square of the 
sum of two numbers. 

3. State a rule for finding automatically the square of the 
difference of two numbers. 

The preceding exercises should establish the following 
short cuts for finding the. square of the sum and the square 
of the difference of two numbers: 

1. The square of the sum of two numbers is equal to the 
square of the first number increased by two times the product 
of the first number and the second number plus the square of 
the second number. Thus, (a -f- ft) 2 = a 2 + 2 ab + b 2 . 

2. The square of the difference of two numbers is equal 
to the square of the first number decreased by two times the 
product of the first number and the second number plus the 
square of the second number. Thus, (a 6) 2 = a 2 2 ab -+- 6 2 . 

EXERCISES 
1. Find automatically the following products : 

(a) (x + 5) (x - 5). (d) (c + 7) (c - 7). 

(b) (a + 4) (a - 4). (e) (y - 11) (y +.11). 

(f) (,. + 8)(r-8). 



2. Study the form of the results in Ex. 1. Give the products 
in the following list at sight : 

(a) (y - 10) (y + 10). (d) (5 -x)(5+ x). 

(b) (2 x + 5) (2 x - 5). (e) (y - J) (// + ). 

(c) (3d-4)(3<* + 4). (f) (5x-y)(5x + y). 

3. Write the sum of x and y ; the difference ; find the prod- 
uct of the sum and the difference. Check by multiplication. 

4. State the rule for finding the product of the sum and the 
difference of two numbers. 



194 GENERAL MATHEMATICS 

The preceding exercises should establish the following 
short cut for finding the product of the sum and the 
difference of two numbers: 

1. Square each of the numbers. 

2. Subtract the second square from the first. 



DRILL EXERCISES 

Find the following products mentally : 

1. (x + 2) (aj + 2). 11. (x - 1) (a; + 1) (* 2 + 1). 

2- (U + 3) (y - 3). 12. (w - c) (w + c) (w* - c 2 ). 

3. (-4)(s-4). 13.- (10 a; + 9) (10 a; + 9). 

4. (2w-5)(2tt; + 6). 14. (y*if - 0.5) (a; 2 / + 0.5). 

5. ( - 2 i) (s + 2 J). 15. (11 +/<7/i 2 ) (11 +/^ 2 )- 

6. (3s + 2o)(3 + 2a). 16. (a 5 + ^ 5 ) (a 5 - 5 ). 

7. (3r-4)(3r + 4). 17. (20 + 2) (20 - 2). 

8 . (Ja + j6)(Ja.-ii). 18. (30 + 1) (30 - 1). 

9- (iy-*)(t*y + *)- 19 - ( 18 )( 22 > 

10. (x - 1) (a + 1). 20. (31) (29). 

243. Division. The law for algebraic division is easily 
learned because of the relation between division and mul- 
tiplication. ^ We recall from arithmetic that division is the 
process of finding one of two numbers when their product 
and the other number are given and also we remember 
that quotient x divisor = dividend. 

These facts suggest the law of division. Thus we know 
that +12-=- + 2 = -f6 because (+ 2)(+ 6)= +12. 



POSITIVE AND NEGATIVE NUMBERS 195 

EXERCISES 

1. Since (2) (- 6) = - 12, what is - 12 -s- 2 ? 

2. Since (- 2) (+ 6) = - 12, what is 12 -s- - 2 ? 

3. Since (- 2) (- 6) = + 12, what is + 12 H- - 2 ? 

4. Since (+ )(+ &) = -f &, what is (+ aV) -+ a? 

5. If the signs of dividend and divisor are alike, what is 
the sign of the quotient ? 

6. If the signs of dividend and divisor are unlike, what is 
the sign of the quotient ? 

244. Law of signs in division. The work of the preced- 
ing article may be summed up in the following law: If 
the dividend and divisor have like signs, the quotient is posi- 
tive ; if the dividend and divisor have unlike signs, the 
quotient is negative. 

245. Dividing a monomial by a monomial. We shall now 
have an opportunity to apply the law learned in the pre- 
ceding article. 

EXERCISES 

Find the quotient in the following, doing mentally as much 
of the work as possible : 

1. (+l5)-(-3)=? 10. (-10ar)'-5-(-2oj)=? 

2. (-15) -L(_3)=? 11- (-/>)* (-)=? 

3. (- 15) -t- (+ 3) = ? 12- (- *) * (+ V) = 'f 

4. (+15)-i-(+3)=? 13- (t *)--(- &) =?- 

5. (- 18) -5- (- 3) = ? 14. (- 0.5 x) + (-$x)= ? 

6 . (-12) -=-(-12)=? 15. (- 1.21 x 2 ) ^ (- 1.1 *) = ? 

7. (+5)-f-(+5)=? 16. (_)-*.(-) = .?. 

8. (+*)^(+*)=? 17. (f)-Kf)=? 

9. -2a-f.a=? 18 ' (-|)^(-f) =? 



196 GENERAL MATHEMATICS 

19 . $).,.(_{).? 31. (-*V(*)=? 

20- (-1) ^ (- 1) = * 32. (- * 2 ) + (^> - 

21. (f)-Kf)='.' 33. (-9^)-(3)=? 

22. (?) + (- t)=- 34 ' (-3*)-l-(-ft)=? 

23. (_2) + (+-J)=? 35. (6fe)-K-2je)= 

36. (+) + (-)=. 



25. (+12./-)-(-x) = ? 38. <_ 

28. (4- ) -*- (- ) = V 39. < t- ww/) -*-( <) = ? 

27. (-*)4-(-J*)=? 40. (-//Ar 8 ) -5-(-oA)='.' 

28. (- .r*) -i- (- x) = ? 41. ( 7 r) -I- (- 22) = V 

29. (^ - (- x) = ? 42. (oi r) -5- (- 3j r) = ? 

30. ( _ x 4 ) -s- (- x) = ? 43. 24 a;// -t- X .? = ? 

XOTK. The algebraic solution of the more difficult problems of 
this type are best interpreted as fractions, since a fraction is an 

_>4 ,- 
indicated quotient. Thus, 24 -/// -=-3 x may be written The 

'^ x 24 j-ii 
problem now is one of reducing to lower terms. Thus, in - 

both numerator and denominator may be divided by 3 x. The result 

8w 

is - > or 8 y units. 

In algebra, as in arithmetic, the quotient is not altered if dividend 
aad divisor are both divided by the same factor. Dividing dividend 
and divisor by the highest common factor reduces the quotient 
(or fraction) to the simplest form (or to lowest terms). 

Solution. The sign of the quotient is negative. Why? 
The numerical factors can be divided l>y S : r 5 and x 1 are divisible 
by x 2 ; y 3 and / s are divisible by y 3 ; m- and w 3 are divisible by ? 2 . 

Hence ***?# = ^ = - f!.- 

8 x-im? m m 



45. 



POSITIVE AND NEGATIVE NUMBERS 197 
-S-jfb 343 



xz 



46.^ 
9 



47. 
48. 



49. 
50. 



49 a; 
12 



246. Dividing a polynomial by a monomial. The division 
process will now be extended. 

EXERCISES 

1. Divide 6 x 2 + 4 xij + 8 xz by 2 a-. 

As in dividing monomials, this quotient may be stated as a rec- 
tangle problem. Find the length of the base of a rectangle whose 
area is 6 x" + 4 xy + 8 xz and whose altitude is 2 x. Indicating this 
quotient in the form of a fraction, 

we have 

6 x z + 4 xy + 8 xz 



3X 



42 



FIG. 168 



Dividing numerator and denominator 
by 2 x, the result is 3x + 2 y + 4~- Show that the problem may now be in- 
terpreted by a rectangle formed by three adjacent rectangles (Fig. 168). 

2. Show that the total area of three adjacent flower beds 
(Fig. 169) may be expressed in either 

of the following forms : 

or 5 (3 + 5 + 4). 

Which form is the better ? Why ? 

3. Find the following quotients, obtaining as many as you 
can mentally : 

9 a 2 -6 a 5 . 27i + 6 2 



5x3 


5x5 


5x4 


354 

Fir.. 169 



(a) 



(g) 



00 



3a 



-4*V 



(e) : 
(f) 



6 r/ 



198 GENERAL MATHEMATICS 

247. Factoring ; prime numbers. To factor a number is 
to find two or more numbers which when multiplied 
together will produce the number. Thus, one may see by 
inspection that 2, 2, and 3 are the factors of 12. 'In like 
manner, x + y and a are the factors of ax 4- ay. 

A number which has no other factors except itself and 
unity is said to be a prime number ; as, 5, x, and a + b. 

A monomial is expressed in terms of its prime factors, 

thus : 1 5 ax*i? = %.5-a-x.x.y.y.y. 

The following is an example of the method of express- 
ing a factored polynomial: 



In algebra, as in arithmetic, certain forms of number 
expression occur very frequently either as multiplications 
or as divisions so much so, that it is of considerable 
advantage to memorize the characteristics of these num- 
bers that we may factor them by inspection and thus be 
able to perform the multiplications and divisions auto- 
matically. In this text we shall study two general types 
of factoring. 

248. Factoring Type I. Taking out a common monomial 
factor. Type form ax + bx + ex = x (a + b + e). 

A number of this type we shall call a number containing 
a common monomial factor. The products obtained in the 
exercises of Art. 239 are numbers of this type. Although 
this type of factoring is not difficult, nevertheless it is im- 
portant and should be kept in mind. We shall learn that 
many verbal problems lead to equations which can readily 
be solved by a method which depends upon factoring. 
Factoring also enables us to transform formulas into their 
most convenient form. 



POSITIVE AND NEGATIVE NUMBERS 199 

The method of factoring this type consists of the 
following steps: 

1. Inspect the terms and discover the factor which is 
common to all the terms. 

2. Divide by the common monomial factor. The result 
obtained is the other factor. 

3. In order to find out whether he has factored correctly 
the student should multiply the two factors together. 

NOTE. In all factoring problems the student should first look to 
see if the number contains a common monomial factor. 

EXERCISES 

Factor the following by inspection and check your work 
by multiplication : 

1 . bx 5 b be. 

Solution. Each term has the factor b. Divide the expression by b. 
The quotient is x 5 c. 

Check. b (x 5 c) = bx 5 b be. 

Therefore the factors of bx 5 6 be are b and x 5 c. 

2. 5a-5b. 8. x* - x 3 . 

3. 4cc + 4?/. 9. 25 x 2 - 5 x 8 . 

4. 5xa-lQxb. 10. 2 x 2 + 4 xy + 2 f. 

5. 5 ax 2 - 10 axif. ' 11. d 2 b + ab' 2 + a 8 . 

6. 2rx 8 -8?y. 12. 4a; 2 -8^ + 47/ 2 . 

7. 3 x 2 - 6 x. 13 . a*a? - 2 aVy 2 + 4 aary . 

14. 3 a 2 -15 a + 18. 

249. Factoring Type II. The " cut and try " method of 
factoring. Type form acx z +(bc+ad')x+bd=:(az+b')(cz+d'~). 

The products obtained in the exercises of Art. 241 can all 
be factored easily by inspection. The method of factoring 



200 GENERAL MATHEMATICS 

such products is known as the "cut and try" or "trial and 
error" method. The method consists simply of guessing 
the correct pair of factors from all of the possible ones 
and then verifying the result by multiplying the factors 
together. The method is illustrated by the following 
example : 

Factor 2 y? + 9 ./ + 10. 

Solution. There are four possible pairs of factors, as shown below : 

2 x + 10 2 x + 10 x + 5 2 x + .5 

x+ 1 2r+ 1 -2x + 2 x + 2 

It is clear that the last pair is the correct one, since the 
sum of the cross-products is 9 x. Of course the correct 
pair of factors may be found at any stage of the "cut 
and try" method, and while the process may seem slow at 
first, practice soon develops such skill that the factors can 
easily be found. 

It is very important for the student always to be sure 
that the factors he has obtained are prime numbers. Such 
factors are called prime factors. Incidentally it is impor- 
tant to remember that there are some numbers that are 
not factorable, because they are already prime numbers. 
For example, a^+16 and 2^ + 2 a; +12 are not factorable. 
See if you can explain why they are not factorable. 

From what has been said the student will see that in 
all factoring problems it is important to hold in mind 
three things; namely: 

1. Try to discover a common monomial factor. 

2. Find the prime factors by the "cut and try" method. 

3. Check by multiplying the factors together. 



POSITIVE AND NEGATIVE NUMBERS 201 

EXERCISES 

Find the prime factors of the following expressions : 
1. 6x*-x-2. 

Solution. Since the x and the 2 are both negative, the last terms 
of the factors are opposite in sign. The possible combinations of 
pairs of factors (regardless of signs) are shown bclov. : 

' Qj; 2 Qx 1 :!./ 2 3x 1 



The third pair is seen to be correct, provided we write them 

8 x - 2 and 2 x + 1. Therefore 6 x 2 - x - 2 - (3 x - 2) (2 x + 1). 

2. 2 a 2 + 9 a + 9. 23. 2 a- 2 - a- - 28. 

3. x 2 + 2 xy + >/ 2 . 24. 5 x 2 - 33 .r + 18. 

4. 2 -16. 25. 16 a- 6 - 25 v/ 4 . 

5. cc 2 + 2x-3. 26. 18 x 2 + 21 x - 15. 

6. x 2 - 2 x - 3. 27. 6 a 2 - a - 2. 

7. 4 a; 2 + 16.x + 16. 28. 16-36a- 6 . 

8. 7 ar + 9 x + 2. 29. a 2 - 4 a - 45. 

9. 9z 2 - 30 a; + 25. 30. 2 x 2 ?/ + 11 ccy + 12 y. 

10. 7/ 4 -s 2 . 31. 2-2x 2 . 

11. x 2 -5a; + 4. 32. 24 #d + 138crf - 36d. 

12. 5 if" 80. 33. (i~ 6 cib 55 & . 

13. 6o; 2 -19x + 15. 34. 2 // 2 - 5 </6 + 6. 

14. y 2 - 5 y - 6. 35. 100 *V - 49. 

15. x 2 -!. 36. 20x 2 -./--99. 
IS. 1 ??i 4 (3 factors). 37. 15 4 j- 4 a- 2 . 

17. 3 a- 2 -IT a- +10. 38. 18 - 33 x + 5 a; 2 . 

18. 2 - 9 2 . 39. 289 W - 81 d*. 

19. 49 -x 2 . 40. 8 a 2 -2. 

20. 3ar 4a;+l. 41. 3 a; 2 + 4 x + 2. 

21. w* a 4- . - 1.2. 42. 19x + 22rK- 2 - 31. 

22. 5 .r 2 -17 x- 12. 43. .r 2 + 20 r + 84. 



202 GENERAL MATHEMATICS 

250. Factoring perfect trinomial squares. Type form 
a? 2a6 + 2 = O ) 2 . Numbers like 4z 2 + 16z + 16 or 
x* 2 xy + y 2 , which are obtained by multiplying a bino- 
mial by itself, are called perfect trinomial squares. They 
are special cases of the second type of factoring discussed 
in Art. 249. We have already seen perfect trinomial squares 
where all the terms are positive in the problems of Ex. 2, 
Art. 127. See if you can formulate a short method of 
factoring perfect trinomial squares. 

EXERCISES 

Factor the following perfect trinomial squares by a short 
method : 

1. a 2 + 2ab + b 2 . 5. 9 x 2 + 42 xy + 49 if. 

2. m 2 -2mn + n 2 . 6. 64 a 2 -32ab + tf. 

3. 9z 2 + 12av/ + 4?/ 2 . 7. 4 x 2 tf - 12 xy + 9 z*. 

4. 16 a 2 - 40 ab + 25 b 2 . 8. 9 aty 4 + 30 afyV + 25 4 . 

251. Factoring the difference of two squares. Type form 
ai &=(a + fr)(a 6). Numbers of the form a 2 b 2 are 
called the difference of two squares. The products obtained 
in the exercises on page 194 are numbers of this type. 
This is a special case of the type discussed in Art. 249. 

ORAL EXERCISES 

1. What is the product of (x + 3) (a; 3)? What then are 
the factors of x 2 9 ? 

2. State the factors of the following : 

(a) x 2 - 4. (c) r 2 - 4 s 2 . 

(b) c 2 -25. (d) 25 a* 

3. Show by means of Fig. 170 on the following page that 

a 2 - b 2 = (a + ft) (a - b). 



POSITIVE AND NEGATIVE NUMBERS 



208 



The equation a 2 b 2 = (a b) (a + &) asserts that a 
binomial which is the difference of two squares may be 
readily factored as follows : M v 

One factor is the sum of 
the square roots of the terms 
of the binomial, and the other 
the difference of the square 
roots of the terms of the 
binomial. 



FIG. 170 



Thus, to factor 49 # 2 6 2 
first find the square root of 
each term ; that is, 7 and ab. Then, according to the rule, 
one factor is 7 + <d> and the other 7 ab. Obviously, the 
factors may be given hi reverse order. Why ? 

EXERCISES 

Factor the following binomials. Check by multiplication 
when you are not absolutely certain the result is correct. 



7. 16 a* -25 6*. 

8. 81rt 2 -16s 2 . 

9. 25 a; 6 -36s 4 . 

10. 49 - 36 a- 6 . 

11. 1- 

12. \ -a- 2 . 



1. a- 2 -16. 
2 ,.-1 ( ,i 

3. //--I. 

4. I-./- 4 . 

5 _ ,2 Q j ,2 

6. 9 - /. 

19. 225 a 6 - wV7t M . 

20. x 4 -y 4 . 

21. 2o?iV- 81m 4 . 

22. a- 8 y 8 . 

23. 625 2 i 4 - 256 a\ 

24. 64 a; 6 -9. 

25. Ca + i) a -9. 



13. 100 4 a; 2 -36. 

14. 289m 2 - 81 ?r. 
15. 

16. 

17. 196 -100 6V. 

18. 361 r 2 ^ 2 - 196. 

26. 9 (a + xf -16. 

27. (x 3 - ?/) 2 - x 6 . 



28. 

29. 0.25 a -0.64J a . 

30. 0.25 r/ 2 -^- 

lo 



204 GENERAL MATHEMATICS 

Knowledge of the special products considered above 
enables us to multiply certain arithmetic numbers with 
great rapidity. Thus the product of 32 by 28 may be 
written (30 + 2) (30 - 2) = (30)2 _ (2 ) = 896. 

EXERCISES 

1. Give mentally the following products : 

(a) 18-22. (e) 32-27. (i) 67-73. (in) 75-85. 

(b)17-23. (f)37-43. (j)66-74. (n) 79 - 81. 

(c) 26 - 34. (g) 38 - 42. (k) 68 . 72. (o) 42 - 38. 

(d)29-31. . (h) 47 -'53. (1)75-65. (p) 95 - 75. 

2. Find the value of the following : 

(a) 71 2 -19 2 . (c) 146 2 -54 2 . (e) 1215 2 - 15 2 . 

(b) 146 2 -46 2 . (d) 312 2 -288 2 . (f) 2146 2 -10 2 . 

252. Different ways of carrying out the same calculations. 
The preceding problems show that the formula a 2 ft 2 
= (a J) (a + J) provides us with a method of making 
calculations easier. In fact, the expressions which are 
linked by the equality sign in a 2 b 2 = (a ) (a + &) 
simply represent two different ways of carrying out the 
same calculations, of which the one on the right is by 
far the easier. 

253. Distinction between identity and equation. An equal- 
ity such as a 2 b 2 = (a 5) (a + />) is called an identity. 
It represents two ways of making the same calculation. 
The statement is true for all values of a and I. The 
pupil should not confuse the meaning of an identity with 
that of an equation. Thus a 2 4 = (x 2) (x + 2) is true 
for all values of x, but y? 4 = 32 is a statement that is 
true only when x = 6 or x = 6 ; that is, it is a state- 
ment of equality in some special situation ; it may be 



POSITIVE AND NEGATIVE NUMBEKS 205 



the translation of an area problem, a motion problem, an 
alloy problem, etc., but it always represents some concrete 
situation, whereas x 2 4 = (x 2) (x + 2) is an abstract 
formula for calculation and is true for all values of x. 

EXERCISES 

1. Tell which of the following are equations and which are 
identities : 

(a) 4 x* - 16 = 20. 

(c) 9z 2 + 12* + 4 = (3z + 2V. _ x 2 -9 

(d) 4 







2. Solve, by factoring, the following equations : 

(a) ax + bx = ac + be. (c) 5 a 2 x 4 b*x = 10 <>' 

ca da 20 

(b)c + f/ = _ + _. r( d > __ s 

/ 6 N 1 2 , ^ 

1 ; 2x-2 3-3" r 4x-4 



6 

,o 
+ 2 



* 254. Calculating areas. The following exercises furnish 
applications of the preceding work of this chapter. 

EXERCISES 

1. Show that the shaded area A in Fig. 171 may be ex- 
pressed as follows: A =(S s)(S + s), where 5 is a side 
of the large square and s a side of the 

small square. 

2. A carpet 20 ft. square is placed in a 
room 25 ft. square. The uncovered border 
strip is to be painted. Find the area of 
the strip. Find the cost of painting this 
area at 80 cents per square yard. Write a 
formula to be used in calculating the cost 
of painting similar strips at c cents per 

yard, the carpet to be x feet square and the room r feet square. 




206 



GENERAL MATHEMATICS 



3. A metal plate is cut as shown in Fig. 172. If 'a =10 and 
b = 2, what is the area of the plate ? In what two ways may 

the calculating be done ? What is the ^ a ^ 

volume of metal if the piece is | in. 
thick ? What is the weight if a cubic 
inch of the metal weighs 20 grams ? 
Write a general formula for a plate 
cut in the form of the figure, t inches 
thick and weighing g grams per 
square inch. Write the result in a 
form which is easily calculated. 



- 




FIG. 173 



... . . FIG. 172 

4. A design pattern is cut in 

the form shown in Fig. 173. Calculate the area. Make a 
verbal problem illustrating this formula. 

5. We can make an application of our 
knowledge of factoring in problems re- 
lated to circles, as will be seen by solving 
the following : 

(a) The area of a circle whose radius 
is r is Trr 2 . What is the area of a circle 
whose radius is R? 

(b) How can you find the area of the ring shaded in 
Fig. 174? Indicate the area. 

(c ) Simplify the result of (b) by remov- 
ing the monomial factor. 

(d) What is the area of a running 
track in which R = 100 and r = 90 ? 

(e) Calculate the area of the shaded 
ring in Fig. 174 if R = 5.5 in. and r = 5 ; 

if R = 3.75 and r = 0.25. 

FIG. 174 

6. Allowing 500 Ib. to a cubic foot, find 

the weight of a steel pipe 20 ft. long if R = 12 in. and r 11 in. 
HINT. Find a rule or formula for the volume of a cylinder. 




POSITIVE AND NEGATIVE NUMBERS 



207 



3C 



FlG 



7. Find the weight of an iron rod 6 ft. long cast in the 

form shown in Fig. 175 if a = 2 in., b \ in y and c = \ in. 

HINT. Allow 500 Ib. per cubic foot. 

255. Division of polynomials 
illustrated by arithmetical num- 
bers. The process of dividing 
one polynomial by another may 
be clearly illustrated by a long- 
division problem in arithmetic ; 

for example, we shall consider 67,942 -r- 322. Ordinarily we 
divide in automatic fashion, adopting many desirable short 
cuts which, though they make our work more efficient, 
nevertheless obscure the meaning. 

In multiplication it was pointed out that because of our decimal 
system the 9 in 67,942 does not stand for 9 units, but for 900 units 
or 9 10 2 units. Similarly, the 7 means 7000, or 7 10 3 , etc. 

If we arrange dividend and divisor in the form of polynomials, 
the division may appear in either of the following forms : 



60000 + 7000 + 900 + 40 + 2 
60000 + 4000 + 400 

3000 + 500 + 40 
3000 + 200 + 20 

300 + 20 + 2 
300 + 20 + 2 



300 + 20 + 2 



200 + 10 + 1 



3-10 2 



10 



6 10* H 


^4-10 3 H 


-4-10 2 


2-10 2 + 10 + 1 




3 10 3 - 
3-10 3 - 


h5-10 2 + 4-10 
h2-10 2 + 2-10 






3 -10 2 + 2-10 + 2 
3 -10 2 + 2 -10 + 2 



The student should study the two preceding examples 
carefully in order to be better able to understand the simi- 
larity of these with the division of algebraic polynomials 
which we shall now discuss. 



208 GENERAL MATHEMATICS 

256. Division of algebraic polynomials. The division of 
algebraic polynomials arranged according to either the 
ascending or the descending power of some letter is similar 
to the preceding division of arithmetical numbers ; thus : 

8 .y 4 + 2 y 3 + 4 y~ 

4 y 3 + 5 y 2 + 3 y 



<r l 



a s + a 2 /; 



a + b 



a?-ab 



- a*b - ab* 

+ aft 2 + b s 
+ ab* + b* 

257. Process in division. From a study of the preceding 
exercises we see that in dividing one polynomial by another 
we proceed as follows: 

1. Arrange both dividend and divisor according to ascend- 
in;! '' descending powers of some common letter. 

2. Divide the first term of the dividend by the first term of 
the divisor and write the result for the first term of the quotient. 

3. Multiply the entire divisor by the first term of the quotient 
and subtract the result from the dividend. 

4. If there is a remainder, consider it as a new dividend 
and proceed as before. 

The student should observe that the process in division 
furnishes an excellent review of the other fundamental 
processes, inasmuch as they are necessary in almost every 
division problem. They should therefore be mastered as 
soon as possible. 



POSITIVE AND NEGATIVE NUMBERS 209 

258. Checking a division. We shall now illustrate the 
method of checking a division : 

Divide x s 3 x 2 y + 3 xy 2 y s by x y. 

X s 3 x 2 y + 3 xy 2 y s \x y 
x 3 x 2 ;/ | x' 2 2 xy -f y 2 

2 x 2 y + 3 xy 1 



xy 2 - y 9 
xy 2 y 3 

First method of checking. Since the division is exact 
(that is, there is no remainder), multiply the divisor by the 
quotient. If the product equals the dividend, the problem 
checks. This may be expressed in cases where there is no 

remainder by the formula q x d = D, or q 

tZ 

How would you check if there were a remainder? If 
the answer is not obvious, try to check similar problems in 
long division in arithmetic. 

Second method of checking. Assume values for x and y. 

Let x = 5 and y 2. 

Substituting in the example, 



and d 3. 

D 27 

Substituting, the formula q = becomes = 9, or 9 = 9. 

d o 

Since 9 (or q) = 9 ( or ), the problem checks. 

\ / 

259. Importance of a thorough drill in division. In the 
process of division practically all the principles of the last 
two chapters are involved. Hence the following exercises 
are important as a means of reviewing the fundamental 
laws of addition, subtraction, multiplication, and division. 



210 GENERAL MATHEMATICS 

EXERCISES 
Divide, and check by either method : 

1. (x 2 - llx + 30) -=-(x- 5). 

2. (// - f - 4y + 4)-<y - 3y + 2). 

3. ( ft s + 7 + 18 a + 40) -r- (a 1 + 2 a + 8). 

4. (9-9x + 8x 2 -4x 8 )-i-(3-2.r). 
5. 

6. (27 x 8 - 54x 2 // + 36xy 2 - 

7. (27 x 8 + 54 afy + 36 xy* + 8 y 8 ) - '(3 x + 2 y). 

8. x 8 --x-7/. 
9. 

10. 

11. (1& + 8 m - 32 m 2 + 32 m s - 15 m 4 ) -=- (3 + 4 m - 5 m 2 ). 

12. (x 8 + 2xy + xz + yz + if) + (x + y + ). 

13. (14x + 2x 4 + llx 2 + 5x 8 - 24)--(2x 2 +'3x - 4). 

14. (r 8 + 65 r - 15 r 2 - 63) -s- (r - 7). 

15. (25 a - 20 a 2 + 6 a 8 - 12) -- (- 4 a + 2 a 2 + 3). 

16. (8x - 4 + 6x 4 + 8x 8 - Ilx 2 )-r-(4x 2 + 2x 3 - x + 2). 

17. (9 x 2 // 2 - 6 x 8 ?/ + x 4 - 4 ?/ 4 ) ^ (x 2 - 3 xy + 

18. (25 x 4 - 60 x 2 / + 36 y 4 ) -=- (5 x 2 - 6 y 2 ). 

19. (4 x 4 + 12 a-y + 9 ?/) H- (2 x 2 + 3 y 2 ). 

20. (a 5 -1) -=-(-!); (a a -l)-s-(a-l). 

21. (a 5 -7/ 5 )-(a-7/); (a*-,ft + (a-y). 
. 22. (25m 4 -49/i 4 )^-(5m 2 + 7w 2 ). 

23. (25 m 4 - 49 w 4 ) -- (5 m 2 - 7 w 2 ). 

24. (0.027 aW + c 8 ) -h (0.2 aft + c). 

25. (8 a 8 - Z> 8 ) -- (4 a 2 + 2 ab + 6 2 ). 



POSITIVE AND NEGATIVE NUMBERS 211 

1 2 3 x 

260. Division by zero. The quotients - - - -, , 

etc. have no meaning, for a number multiplied by 
cannot give 1, 2, 3, x, etc. (see the definition of division in 

Art. 243). The quotient - is undetermined, as every num- 
ber multiplied by equals 0. Therefore we shall assume 
that in all quotients hereafter the divisor is not zero nor 
equal to zero. 

EXERCISES 

1. The following solution is one of several that are some- 
times given to show that 1 = 2. Find the fallacy. 

Two numbers are given equal, as x y. 

Then x - y - 0, Why ? 

and 2 (a; - y) = 0. Why ? 

Then x - y = 2 (x - y). Why ? 

Dividing both sides by x y, 1 = 2. 

2. Give a similar argument which seems to show that 2 
equals 5. 

SUMMARY 

261. This chapter has taught the meaning of the follow- 
ing words and phrases : turning tendency, force, lever arm, 
multiplication, division, factoring, factors, prime number, 
number containing a monomial factor. 

262. The law of signs in multiplication was illustrated 
(1) geometrically with line segments and (2) by means 
of the balanced beam. 

263. The following agreements were used : 

1. A weight pulling downward is negative; one pulling 
upward is positive. 

2. A force tending to rotate a bar clockwise is negative ; 
counterclockwise, positive. 



212 GENERAL MATHEMATICS 

3. A lever arm to the right of the point where the bar is 
balanced is positive ; to the left, negative. 

The turning tendency (or force) acting upon a balanced 
bar is equal to the product of the weight times the lever arm. 

264. Law of signs in multiplication: The product of 
two numbers having like signs is positive ; the product of two 
numbers having unlike signs is negative. 

265. The chapter has taught and geometrically illus- 
trated the following processes of multiplication: 

1. The multiplication of several monomials. 

2. The multiplication of a monomial by a polynomial. 

3. The multiplication of polynomials by polynomials. 

The order of factors may be changed without changing 
the product. 

The value of a product is zero if one of the factors 
is zero. 

266. Law of division : The quotient of two numbers hav- 
ing like signs is positive ; the quotient of two numbers having 
unlike signs is negative. 

Arithmetical numbers may be arranged in the form of 
polynomials according to powers of 10. 

The process of dividing one polynomial by another is essentially 
the same as the process of dividing arithmetical numbers. 
In all problems of the text the divisor is not zero. 

267. The chapter has taught the following forms of 
division : 

1. The division of a monomial by a monomial. 

2. The division of a polynomial by a monomial. 

3. The reduction of a fraction to lowest terms. 

4. The division of a polynomial by a polynomial. 



POSITIVE AND NEGATIVE NUMBERS 213 

268. Division has been illustrated geometrically. Two 
methods for checking division were taught. 

269. The following types of factoring were taught: 
Type I. Taking out a common monomial factor, 

ax + bx -f ex = x (a + b + c). 
Type II. The " cut and try " method, 

acx 2 + (be + ad~) x+bd = (ax + 6) (ex + d). 



CHAPTER X 



GRAPHICAL REPRESENTATION OF STATISTICS; THE 
GRAPH OF A LINEAR EQUATION 

270. Facts presented in the form of a table. The follow, 
ing table of facts shows in part the recreational interests 
of the boys and girls of certain Cleveland (Ohio) high 
schools. Thus, of 4528 boys, 4075 play baseball ; of 3727 
girls, 2608 play baseball ; 7402 children out of a total of 
8255 attend the movies regularly; and so on. 

TABLE OF RECREATIONAL INTERESTS 





BOYS 


GIRLS 


TOTAL 


Number of students 


4528 


3727 


8255 


Number who play baseball 


4075 


2608 


6683 


Number who play basketball 
Number who play tennis 


3018 
1811 


1390 
1864 


4408 
3675 


Number who belong to Camp Fire Girls . 
Number who wrestle .... 


1358 


621 


621 
1358 


Number who attend movies 


4010 


3392 


7402 


Number who attend movies daily .... 


754 


485 


1239 



EXERCISE 

Study the preceding table until you understand the mean- 
ing of the columns of figures. 

271. Pictograms; graphs. Tables made up of columns of 
figures are common in newspapers, magazines, and books, 
but a table like the preceding is not the best device for 

214 



REPRESENTATION OF STATISTICS 215 

expressing the meaning of an array of facts. The ordinary 
mind cannot see the relations if the list is at all extended ; 
hence it often happens that the real meaning of a series of 
facts is lost in a complicated table. Newspapers, magazines, 
trade journals, and books, realizing this fact, are beginning 
to add to tables of statistics pictures which show their 
meaning and their relationships more clearly than can be 
done by columns of figures. 

The significance of the facts of the preceding table is 
far more vividly expressed by the pictures of Fig. 176. 
Thus the pictures show that of the high-school girls one 
out of every two (50 per cent) plays tennis ; two out of 
every dozen (16| per cent) are Camp Fire Girls ; of the 
high-school boys six out of every twenty (30 per cent) 
wrestle ; 85 per cent of all the elementary-school and 
high-school boys attend the movies regularly; and so on. 

The pictures constitute a more powerful method of 
teaching numerical relations, because they teach through 
the eye. For this reason they are called graphic pictures, 
pictograms, or simply graphs. 

EXERCISES 

By means of the pictograms in Fig. 176 answer the following 
questions : 

1. What per cent of the Cleveland boys play tennis ? of 
the Cleveland girls ? With which group is tennis the more 
popular ? 

2. Assuming that every sound-bodied boy should learn to 
wrestle, does your class make a better or a worse showing in 
per cents than the Cleveland boys ? 

3. Are a larger per cent of the girls of your class Camp 
Fire Girls than is the case in the Cleveland high schools ? 




High-School Girls 
High-School Boys 
High-School Girls 
High-School Boys 
High-School Girls 



Do not play Tennis 
Do not play Tennis 



We Never Do 



We Never Do. 



We Never Do 




We play Baseball I Never Do 

o^o High-School Girls ^^^^^^U^AA, 

Belong to Camp Fire Do not belong to Camp Fire 



High-School Boys 



Wrestle Do not Wrestle 

Elementary- and High-School Boys 



Attend Movies 



Daily Nonattendance 
High-School Boys \j 

Help! 

High-School Girls ., laU_^W^_ 
S. 0. S, Board of Education ! 




Do not attend Movies 



FIG. 176. SHOWING HOW PICTOGRAMS ARE CSED TO EXPRESS FACTS 

(Adapted from Johnson's "Education through Recreation ") 

216 



REPRESENTATION OF STATISTICS 



217 



4. Compare your class's swimming record with the Cleve- 
land record. 

5. Continue the discussion with your classmates until it is 
clear to you just how the figures of the pictograms represent 
the facts of the table. 

272. The circle pictogram. The circle is frequently used 
to show quantitative relations. It shows two things: 

(1) the relation of each magnitude to each of the others ; 

(2) the relation of each magnitude to the sum of all. 
The graph in Fig. 177 shows that in a certain year 

Portland (Oregon) spent 30.7$ out of every dollar on 
its school system. This is prac- 
tically j 3 ^ of all the money spent 
by the city and about 1^ times 
as much as the next highest 
item. Compare the part spent 
for education with each of 
the other parts shown in the 
figure. 

The scale used in making 
circle pictograms is based on 
the degree. The angular space 
around the center of the circle 
(360) is divided into parts so 
as to express the numerical re- 
lations ; for example, since almost ^ of the money is 
spent for education, an angle of ^ of 360, or 108, is 
constructed with the protractor at the center of the circle 
and the sides of the angle extended until they intersect 
the circle. The sector (the part of a circle bounded by 
two radii and an arc) formed shows the part of the city's 
money that goes toward education. 




FIG. 177. THE CIRCLE 
PICTOGRAM 

(From Cubberly's " Portland 
Survey ") 



218 (iKXERAL MATHEMATICS 

EXERCISES 

1. In Fig. 177 how does the amount paid as interest com- 
pare with the amount paid for police and fire protection ? 

2. Do you think it would have been more profitable in the 
long run for Portland to pay cash for all public improvements ? 

HINT. A definite answer to this problem may be obtained if 
several members of the class will solve Ex. 3 and report to the class. 

*3. What will it cost a city to build a $100,000 high-school 
building if $20,000 of its cost is paid in cash and the remainder 
paid by issuing 4 per cent bonds of which $4000 worth are to be 
retired (paid) annually ? (All interest due to be paid annually.) 
NOTE. The problem must not be interpreted as an argument 
showing that bonding (borrowing) is never justifiable. 

*4. The discussion of the method of paying the expenses 
of the United States for the first year of our participation in 
the European War was sharply divided between two groups. 
One group favored a large amount of borrowing by the issu- 
ance of bonds, while the other advocated a pay-as-you-go 
policy, that is, raising the money by taxation. Debate the 
merits of the two plans. 

5. Show the following facts by means of a circle divided 
into sectors : 

TABLE SHOWING DISPOSITION OF THE GROSS REVENUE OF 
THE BELL TELEPHONE SYSTEM FOR THE YEAR 1917 
ITEMS PER CENT 

Salaries, wages, and incidentals 50 

Taxes . 7 

Surplus 4 

Materials, rent, and traveling expenses .... 20 

Interest 7 

Dividends 12 

Though widely used, the circle divided into sectors is 
not a quite satisfactory method of showing the ratio of 



REPKESENTATION OF STATISTICS 



219 





numbers. In fact, the objections are so serious that the 
method of construction was given to protect the student 
against false conclusions. The method is not inaccurate 
when the parts which constitute a unit are shown by the 
use of one circle. It frequently 
happens, however, that the com- 
parison is made by circles differ- 
ing in size. In such a case, since 
the eye tends to make the com- 
parison on an area basis, the ratio FlG - 178 - CIRCLES DRAWN ON 

P.I v 1111 AN AREA BASIS SHOWING THE 

of the two numbers should be ex- BB op BANK DEPOSITORS 

pressed by the ratio of the areas 

of the two circles, and statistical authorities so recommend. 
In Fig. 1 78 the circles are drawn on an area basis, but the 
right-hand circle appears less prominent than the figures 
justify, thus causing the reader 
to underestimate the ratio. In 
Fig. 179 the circles are drawn 
on a diameter basis. The right- 
hand circle appears more prom- 
inent than the figures justify, 
thus causing the reader to over- 
estimate the ratio. This feature 
is frequently utilized by those who make dishonest use 
of circle diagrams. The conclusion is that a comparison 
between circles differing in size should be avoided alto- 
gether. Better graphic methods will be taught. Space is 
given here to circle pictograms because of their extensive 
use in many fields. 

EXERCISE 

Test the accuracy of circle pictograms which you may find 
in magazine articles and advertisements. Discuss their value 
with your classmates. 




FIG. 179. CIRCLES DRAWN ON 
A DIAMETER BASIS 



220 



GENERAL MATHEMATICS 



1911 



1899 



273. Area pictograms. The picture of the 
two traveling men given here is intended to 
show the increase in the passenger traffic of 
the railroads. The two men are compared 
on the basis of height. 
The 1911 man, on 
account of his far 
greater area, looks 
more than 2^ times 
as large as the 1899 
man. The men should 
be compared on the 
basis of area. 1 This 
type too should be 
avoided because it tends to deceive the ordinary reader. 





14,591,000 One Mile 32,837,000 One Mile 



FIG. 180. A POPULAR TYPE OF PICTOGRAM, 
TO BE AVOIDED 



1911 



32.837000 

ONE MILE 



mmmmM 



1899 



14.59I.OOO 
ONE MILE 




FIG. 181. A MORE ACCURATE METHOD OF PORTRAYING FACTS 

EXERCISES 

1. Why would it be difficult to make a drawing on the 
basis of area? 

2. Do you know any method which could be used to check 
a drawing made on the basis of area ? (See Art. 109.) 

274. Volume or block pictograms. Cubes, parallelepipeds, 
and spheres are frequently used in comparing relative 
volumes ; for example, pictures of bales of hay or cotton 

1 Brinton, in his excellent text, ' ' Graphic Methods for Presenting Facts,' ' 
presents a chart (Fig. 181) drawn from the same facts as that in Fig. 180. 
Note that the facts are portrayed much more clearly and accurately. 



REPRESENTATION OF STATISTICS 221 

are used to show the output of the states producing these 
articles. The comparison should be made on a basis of 
volume, but often there is no way for the reader to tell 
on what basis the drawing was constructed, whether by 
height, area, or volume. Certainly it would be difficult to 
check the statement made in such a case. 

275. Limitations of area and volume pictograms. The 
student will need to remember that in a correctly con- 
structed area graph the quantities represented should vary 
directly as the number of square units within the out- 
lines of the figures. Thus, in the comparison of passenger 
service relative size should not be determined by the 
relative heights of the men but by the number of square 
units within the outlines. Hence a rough method of 
checking is to transfer the pictures of the traveling men to 
squaredpaper by means of tracing paper and compare the 
number of square millimeters in the area of each with the 
corresponding facts of the table. Similarly, in accurate 
volume or block graphs the quantities should vary as the 
number of cubic units. 

Many who use this form of statistical interpretation 
carelessly fail to observe these principles, and the diffi- 
culty of a check makes this form of graph a convenient 
device for those who would dishonestly misrepresent the 
facts. The general public is not always able to interpret 
the graphs correctly even if they have been properly drawn. 
Because of these limitations it is somewhat unfortunate 
that this type of graph is so extensively used in bulletins 
and current magazines. 

EXERCISE 

Try to obtain and present to the class an advertisement 
illustrating the misuse of an area or volume pictogram. 



222 



GENERAL MATHEMATICS 



276. Practice in interpreting the bar diagram. Fig. 182 
shows one of the suggestions of the Joint Committee on 
Standards for Graphic Presentation. The diagram, Fig. 182, 
(a), based on linear measurement, is called a bar diagram. 
We shall study this topic further in the next article. Review 



Year 

1900 



1914 



FIG 



Tons 

270.588 





555,031 

(a) (b) 

182. BAR DIAGRAMS SHOW FACTS BETTER THAN AREA AND 
VOLUME PICTOGRAMS 



the objections to the other two diagrams (the squares and 
blocks shown in Fig. 182, (b)). Where it is possible the 
student should represent quantities by linear magnitudes, 
as representation by areas or volumes is more likely to 
be misinterpreted. 

EXERCISES 

1. Study Fig. 183 and determine to what extent the two 
horizontal bars are helpful in expressing the ratio of the two 
numbers given. 

2. Would the UXfl 1 Cotton, $820,320,000 
bars in Fig. 183 

be sufficient with- 
out the illustra- 1HI" Wheat, $561,051,000 
tions at the left 
of the numbers ? ^ IG ' ^' ^ F AIR DIAGRAM. (AFTER BRINTON) 

3. With the aid of compasses check the accuracy of Fig. 184. 
Note that the figures are written to the left of the bars. In many 

woe $4.409,136 ^1 bar diagrams the figures 

^^^^^^^^^^ are written to the right 

1912 28 soo 139 ^^^^m ^ ^ e bars. Can you 

FIG. 184.1 DIAGRAM SHOWJNG .N Ex- think of a serious ob .l ec - 

PORTS OF AUTOMOBILES. (AFTER BRINTON) tion to that method ? 

1 See paragraph 7 under Art. 277 




REPRESENTATION OF STATISTICS 



223 



4. Why is there a space left between the bars for 1906 and 
1911 in Fig. 184 ? Do you see any other way to improve the 
diagram ? (See Art. 277.) 

5. Draw on the blackboard a figure similar to Fig. 184, 
adding a bar for the 

year 1917. (The sum 
for this year is about 
900,000,000.) 

6. Explain Fig. 185. 

7. Show that the 
bars of Fig. 186 reveal 

more clearly than the following table the rank of the United 
States in respect to wealth. These are the 1914 estimates. 

United States $150,000,000,000 




FIG. 185. DIAGRAM SHOWING DEATH 
KATE FROM TYPHOID IN 1912 PER HUN- 
DRED THOUSAND POPULATION 



Great Britain and Ireland 

Germany 

France 

Russia 

Austria-Hungary . . . 
Italy 



85,000,000,000 
80,000,000,000 
50,000,000,000 
40,000,000,000 
25,000,000,000 
20,000,000,000 



8. Show that it would have been as accurate and more con- 
venient to draw the preceding diagrams on squared paper. 



75 100 125 150 



UNITED STATES 150 

GREAT BRITAIN or 
AND IRELAND- 5 

GERMANY 80 
FRANCE - - 60 


! <; ' f j 
















i ; - 1 ' \ ' 




';>>' \ ; 


: . U 














RUSSIA -.40 








' -- i 


AUSTRIA-HUNGARY. 25 
ITALY 20 






. . .;>; .. 


mm 





FIG. 186. COMPARATIVE WEALTH OF NATIONS IN 1914 

9. The table for the wealth of nations contains estimates 
prepared at the beginning of the European War (1914-1918). 



-2'24 - GENERAL MATHEMATICS 

These estimates are now far from facts. The student should 
attempt to get the latest estimates from the " World Almanac " 
and construct a bar diagram which will present the situation to 
date and will enable him to make an interesting comparison. 

10. Discuss bar diagrams similar to those given on pages 
222-223 which you may find in Popular Mechanics Magazine, 
Motor, Popular Science Monthly, and Industrial Management. 
For the time being limit yourself to the simpler diagrams. 

277. How to construct a bar diagram. An understanding 
of how to construct bar diagrams and how to interpret 
those he may find in newspapers and magazines should 
be a part of the education of every general reader, just as 
it is of every engineer, physician, statistician, and biologist. 1 
As civilization advances there is being brought to the atten- 
tion of the reading public a constantly increasing amount 
of comparative figures of a scientific, technical, and statis- 
tical nature. A picture or a diagram which presents such 
data in a way to save time and also to gain clearness is 
a graph. The bar diagram is a widely used method of con- 
veying statistical information graphically. The solution of 
the introductory exercises along with the discussion of such 
supplementary graphs as may have seemed profitable for the 
class to discuss will help the pupil to understand the follow- 
ing outline of the method of constructing a bar diagram : 

1. The bars should be constructed to scale. To obtain 
a convenient unit first inspect the size of the smallest 
and the largest number and then choose a line segment to 

1 Neither pupils nor teachers should be misled by the apparent sim- 
plicity of this work. The details are of the greatest importance. It will 
be helpful to obtain the reports of the Joint Committee on Standards for 
Graphic Presentation. This is a competent committee of seventeen, which 
has expended considerable effort on these elementary phases. The pre- 
liminary report may be had from the American Society of Mechanical 
Engineers, 29 W. 39th St., New York ; price, 10 cents. 



KEPRESENTATION OF STATISTICS 225 

represent a number such that it will be possible to draw an 
accurate bar for the smallest number and a bar not too long 
for the largest number. The lines in Fig. 186 on page 223 
are so constructed that the relation between the lengths 
of any two is the same as the relative size of the quan- 
tities represented. A line segment 1 mm. long represents 
$2,500,000,000 of wealth. In the table on page 223 the 
United States is- estimated as possessing three times as 
much wealth as France, and so the line segments repre- 
senting the wealth of the United States and France are 
respectively 60 mm. and 20 mm. long. 

2. The scale and sufficient data should appear on the 
diagram. 

3. Each bar should be designated. 

4. The bars should be uniform in width. 

5. The diagram, should have a title or legend. 

6. Accuracy is the important characteristic. 

7. The space between the bars should be the same as the 
width of the bars, except in a case like Fig. 184, where a 
larger space indicates that the three bars do -not represent 
consecutive years. 

8. In general the zero of the scale should be shown. How- 
ever, there are exceptions ; for example, in graphing the tem- 
perature of a patient we are particularly concerned with how 
much above or below the normal the patient's temperature 
is. Hence, in a case like this we should emphasize the 
normal temperature line. 

EXERCISES 

1. Present the facts of the table given on page 226 by means 
of a bar diagram, using the scale 1 mm. = f 2,000,000. The table 
is arranged to show the twenty heaviest buyers of American 
goods, as indicated by the value of exports from the United 
States during the fiscal year 1914. 



GE;NEKAL MATHEMATICS 



AMERICA'S TWENTY BEST CUSTOMERS 
(From the report of the Bureau of Foreign and Domestic Commerce) 





VALUE OF 
PURCHASES 




VALUE OF. 
PURCHASES 


1. England . . 


$548,641,399 


11. Argentina . . 


$45,179,089 


2. Germany . 


344,794,276 


12. Mexico 


38,748,793 


3. Canada . . . 


344,716,981 


13. Scotland . . . 


33,950,947 


4. France . . . 


159,818,924 


14. Spain . . . . 


30,387,569 


5. Netherlands . 


112,215,673 


15. Russia .... 


30,088,643 


6. Oceania . . 


83,568,417 


16. Brazil .... 


29,963,914 


7. Italy . . . 


74,235,012 


17. China .... 


24,698,734 


8. Cuba . . . 


68,884,428 


18. Austria-Hungary 


22,718,258 


9. Belgium . . 


61,219,894 


19. Panama . . . 


22,678,234 


10. Japan . . . 


51,205,520 


20. Chile .... 


17,482,392 



*2. If possible, ascertain the facts to date (see "World Alma- 
nac "), graph results as in Ex. 1, and compare the two diagrams. 
Account for unusual changes. Are new customers appearing 
among the " twenty best " ? Have old ones dropped out ? 

3. Present the statistics of the following table by means of a 
bar diagram showing the comparative length of rivers. (Use the 
scale 1 cm. = 400 mi. ; the lengths given in the table are in miles.) 





LENGTH 




LENGTH 


Missouri-Mississippi . . 


4200 


Volga 


2400 


Amazon 


3800 


Mackenzie 


2300 


Nile 


3766 


Plata . . ... 


2300 


Yangtze 


3400 


St Lawrence 


2150 


Yenisei 


3300 


Danube 


1725 


Kongo 


3000 


Euphrates 


1700 


Lena 


3000 


Indus 


1700 


Niger 


3000 


Brahmaputra 


1680 


Ob* 


2700 


Ganges 


1500 


Hoangho 


2600 


Mekon^ 


1500 


Amur 


2500 


Rio Theodoro 


950 











REPRESENTATION OF STATISTICS 



227 



4. Represent the statistics of the following table by bar 
diagrams. The estimates of the leading crops in the United 
States for the year 1917 are here compared with the revised 
figures for the crops of the preceding nine years. The pupil 
should note that each column is a separate problem. 



REPORT OF THE UNITED STATES DEPARTMENT OF 
AGRICULTURE FOR 1917 



YEAR 


CORN 


WHEAT 


OATS 


BARLEY 


RYE 


COTTON 




Bushels 


Bushels 


Bushels 


Bushels 


Bushels 


Bales . ' 


1917 .... 


3,159,494,000 


650,828,000 


1,587,286,000 


208,975,000 


60,145,000 


10,949,000 


1916 .... 


2,583,241,000 


639,886,000 


1,251,992,000 


180,927,000 


47,383,000 


12,900,000 


1915 .... 


2,994,793,000 


1,025,801,000 


1,549,030,000 


228,851,000 


54,000,050 


12,862,000 


1914 .... 


2,672,804,000 


891,017,000 


1,141,060,000 


194,953,000 


42,778,000 


15,136,000 


1913 .... 


2,446,988,000 


763,380,000 


1,121,768,000 


178,189,000 


41,381,000 


14,552,000 


1912 .... 


3,124,746,000 


730,267,000 


1,418,377,000 


223,824,000 


35,664,000 


14,104,000 


1911 .... 


2,531,488,000 


621,338,000 


922,298,000 


160,240,000 


33,119,000 


16,101,000 


1910 .... 


2,886,260,000 


635,121,000 


1,186,341,000 


173,832,000 


34,897,000 


12,075,000 


1909 .... 


2,552,190,000 


683,350,000 


1,007,129,000 


173,321,000 


29,520,000 


10,513,000 


1908 .... 


2,668,651,000 


664,002,000 


807,156,000 


166,756,000 


31,851,000 


13,817,000 



278. Bar diagrams used to show several factors. We 
shall now see how bar diagrams may be used to show 
several factors. 

INTRODUCTORY EXERCISES 

1. Fig. 187, on page 228, differs from those in Art. 277 
in that it presents two factors. What is the scale of the dia- 
gram ? Note that the bars representing new buildings extend 
from the top to the bottom of the black bar. Try to account for 
the heavy losses by fire in 1904 and 1906. Why is the bar so 
short for new buildings in 1908 ? (The values are given in 
millions of dollars.) Criticize this diagram according to the 
principles of Art. 277. 

2. Give the facts of Fig. 187 for the twelfth year; the 
eighteenth year. 



228 



GENERAL MATHEMATICS 



JNew Building 



Fire Losses 




1901 1902 1903 1904 



1905 



1906 1907 1908 1909 1910 1911 



FIG. 187. DIAGRAM OF YEARLY VALUES OF NEW BUILDINGS, AND OF ALL 
BUILDINGS LOST BY FIRE IN THE UNITED STATES, 1901-1911, INCLUSIVE 

(Courtesy of W. C. Brinton) 

3. Fig. 188 shows the business relations involved when a city 
bonds itself to buy some present need or luxury. The parts 
of a single bar (say the tenth) show the following: (a) the 
interest paid to date (the black portion) ; (b) the amount 
of the $75,000 still outstanding (the plain portion) ; (c) the 
part of the debt that has been paid (the crosshatched portion). 

Show that a public bond issue is not only a debt but that 
it " conies dangerously near to a perpetual tax," 



REPRESENTATION OF STATISTICS 



229 



19 20 




100.000 
90.000 
80,000 
70,000 
60.000 
50.000 
40,000 
30.000 
20,000 
10.000 



"- 

FIG. 188. BAR DIAGRAM USED TO SHOW MONEY TRANSACTIONS 
INVOLVED IN PAVING FOR A $75,000 SCHOOL BUILDING 

(Adapted from Ayres's " Springfield Survey ") 

The preceding exercises show how a bar diagram may 
be used to compare several factors of some problem 
which are more or less related. If the pupil is especially 
interested in this side of the subject, he may do the 
following exercises. The topic is not particularly impor- 
tant, however, because another method which we shall 
presently stftdy is much more efficient. 

EXERCISES 

*1. Go to the township, county, or city-hall authorities and 
find out how one or more of your public improvements is 
being paid for ; that is, find out (a) if bonds were issued ; 

(b) how many dollars' worth are retired (paid for) each year ; 

(c) how much interest must be paid each year. Construct a 



230 GEXEKAL MATHEMATICS 

bar diagram similar to the one reprinted from the Ayres report, 
showing what it will ultimately cost your community to pay 
for the project. 

*2. A certain county in Indiana built 20 mi. of macadam- 
ized road by issuing $40,000 worth of 4 per cent nontaxable 
bonds. Twenty $100 bonds were to be retired each year. By 
means of a bar diagram show how much it ultimately cost this 
township to build its turnpike. 

*3. Ten years after the turnpike referred to in Ex. 2 was 
built it was practically worn out. Did this township lend to 
posterity or borrow from it ? Give reasons for your answer. 

279. Cartograms. Statistical maps which show quanti- 
ties that vary with different geographic regions are some- 
times called cartograms. The student will doubtless find 
examples in his geography. He should also examine the 
" Statistical Atlas of the United States," published by 
the Census Bureau. Various colors and shades are used 
to help express the meaning. 

When the cost of color printing is prohibitive the same 
ends may be attained by Crosshatch work. The student 
should examine rainfall maps containing cartograms and 
which are often printed in newspapers. 

A special form of cartogram is the dotted map. If \ve 
wish to show the density of population of a city, we may 
take a map of-the city and place a dot within a square 
for every fifty people living in the square. The scale 
should be so chosen that the dots will be fairly close 
together in the sections whose population is of greatest 
density. Space is not given here to illustrating this type, 
but the pupil will have no difficulty with the exercises 
that follow. 



REPKESENTATION OF STATISTICS 



231 



EXERCISES 

1. Obtain at least five different forms of pictograms and 
cartograms from newspapers, magazines, trade journals, or 
government bulletins. Explain very briefly what each intends 
to show. 

2. Discuss the merits or defects of the graphs of Ex. 1. 

280. Interpreting (or reading) graphic curves. The intro- 
ductory exercises given below will furnish the student with 
some practice in the interpretation of graphic curves. 



1915- 




-1916- 



\7_ 



-1917- 



FIG. 189. SHOWING RAILWAY-STOCK FLUCTUATIONS, BY MONTHS, IN 

THE AVERAGE PRICE OF TWENTY-FIVE OF THE LEADING STOCKS ON 

THE NEW YORK STOCK EXCHANGE 

(Adapted from the New York Times) 



INTRODUCTORY EXERCISES 



1. Explain the curve in Fig. 189, noting the highest price, 
the lowest price, the cause of the upward trend in 1915, the 
cause of the downward movement in 1917, and the cause of 
the sharp break upward in the closing days of 1917. 



232 



GENERAL MATHEMATICS 



2. Explain the curve in 
Fig. 190. Check the graph 
for the early years. Give a 
reason for such results as you 
may find. 

3. Fig. 191 is a temperature 
chart of a case of typhoid fever, 
(a) Explain the rise and fall 
of the curve, (b) What is the 
meaning of the dots ? Do these 
points mark the tops of bars ? 

(c) What assumption does the 
physician make when he con- 
nects these paints by a curve ? 

(d) Note that this diagram does 
not have a zero scale ; why was 
it omitted? The chart would 
be improved if it had an empha- 



1910 1911 1912 1913 1914 1915 1916 1917 1918 
90 



(40 




4.3 



FIG. 190. NUMBER OF CARS OF 
DIFFERENT TYPES 

(Adapted from Motor) 



sized line representing normal temperature (98.4). Why ? Con- 
struct a line in color in your text for the normal-temperature line. 



/7 



/jajt&t 




V& 



II IS 



It/7 



107* 
106' 
105 
104* 
103 
102* 
101" 

100' 

? 

"Itf 



YU590 bl 



&S> 1 



Si/eitr-va 



FIG. 191. A "TEMPERATrRE CHART OF A CASE OF TYPHOID FEVER 



KEPKESENTATIOX OF STATISTICS 



233 



4. Explain the heavy line (temperature) of Fig. 192. 

5. Explain the light line (wind) of Fig. 192. 

6. If your study of science has familiarized you with the 
term " relative humidity," explain the dotted line of Fig. 192. 



SPCCSPCSPCSPCCRSSCS PCPC RPCRPCRR CRRRRR PCPC 




2 4 6 8 10 12 14 16 18 20 22 24 26 28 80 

FIG. 192. WEATHER RECORD IN ST. Louis FOR OCTOBER, 1917 

The heavy lines indicate temperature in degrees F. The light lines indicate 
wind in miles per hour. The broken lines indicate relative humidity in percen- 
tage from readings taken at 8 A.M. and 8 P.M. The arrows fly with the pre- 
vailing direction of the wind. S, clear ; PC, partly cloudy ; C, cloudy ; R, rain. 
(From the Healing and Ventilating Magazine, December, 1917) 

The curves of the preceding exercises are called graphic 
curves. The graphic curve is particularly useful in com- 
paring the relation that exists between ttvo quantities ; for 
example, the relation between the price of wheat and the 
annual yield ; the relation between the price of beef and 
the amount produced or exported; the relation between 
office expenses and the size of the corporation ; the tariff 
necessary in order that an article may be manufactured in 
this country ; and so on. 

281. How the graphic curve is drawn. The graphic curve 
in Fig. 193 represents the growth of the population in the 
United States from 1790 to 1910, as shown by the table 
on page 234. 

(1) The horizontal line OM represents the time line. 
1 mm. represents 2 yr. ; that is, 10 yr. is represented by 



234 



GENERAL MATHEMATICS 



YEAK 


POPULATION 


YEAR 


POPULATION 


1790 . . . 


3,929,214 


I860 


31 443 321 


1800 


5,308,483 


1870 


38 558 371 


1810 


7 239 881 


1880 


50 155 783 


1820 


9,638 453 


1890 


62 947 714 


1830 


12,860,702 


1900 


75 994 575 


1840 


1 7 063 363 


1910 


91 972 266 


I860 


23,191,876 















two small spaces. (2) The vertical scale represents the 
population in millions. Two small spaces represent ten 
million. Therefore a bar about 1.6 mm. long is placed on the 
1790 line, and a second bar a little over 2 mm. long is placed 
on the 1800 line. 
Similarly, bars 
were placed on 
the other lines. 
(3)The upper end 
points of the ver- 
tical segments 
(bars) are joined 
by a curve. In 
so far as the 
bars are con- 
cerned the fig- 
ure does not 
differ from an 
ordinary bar diagram. We may assume, however, that in- 
crease in population between any two periods was gradual; 
for example, we may estimate that in 1795 the popula- 
tion was reasonably near some number halfway between 
3,900,000 and 5,300,000 ; that is, about 4,600,000. Simi- 
larly, we may estimate the population in the year 1793. 



upr 
















































































































so 






















































y 


J n 
























































1 




















































/ 




I 


















































/ 






















































f 
























































/ 








CO 














































/ 










rn 












































J/ 






















































x 






















































x* 






















































X 




















































- 


^ 




















































x 




















































^ 


X 


















































X 


X 


















































* 


^ 


















































r^ 
















































- ' 















































































































r 




1 


0- 


-J 


O- 


| 


s 


J 




_c 


S- 


u 

-0 


5 
I- 


J 


R- 


j 


n- 



-0 



0- 


-j 




:\ 


. 


-35-. 



















































































































FIG. 193. THE GRAPHIC CURVE 



REPRESENTATION OF STATISTICS 



235 



This assumption leads us to draw the smooth curve which 
enables us to estimate the change in population without 
knowing the exact length of the bars. By means of the 
curve predict what the population will be in 1920. In what 
way will the accuracy of your prediction be affected by the 
European War? 

EXERCISES 

1. The following table shows the total monthly sales of a 
bookstore through a period of two years. 



January .... 
February .... 


1917 
$2125 
2237 


1918 

$2329 
2416 


July 
August 


1917 

$2271 
2231 


1918 
$2380 
2350 


March 


2460 


2479 


September .... 


2542 


2620 


April 


2521 


2590 


October ... 


2725 


2831 


May . 


2486 


2580 


November .... 


2345 


2540 


June 


2393 


2482 


December . 


2825 


3129 















Draw a graphic curve for each year on the same sheet of 
graphic paper. Draw the two curves with different-colored 
ink or else use a dotted line for one and an unbroken line for 
the other. Explain the curves. 

2. On a winter day the thermometer was read at 9A.M. and 
every hour afterward until 9 P. ai. The hourly readings were 
- 5, 0, - 2, - 8, - 10, - 10, - 5, 0, - 5, - 4, - 2, 
3, 7. Draw the temperature graph. 

3. Using a Convenient scale and calling the vertical lines 
age -lines, graph these average heights of boys and girls : 



AGE 


BOYS 


GIRLS 


AGE 


BOYS 


GIRLS 


2yr. 


1.6ft. 


1.6ft. 


12 yr. 


4.8ft. 


4.5ft. 


4 


2.6 


2.6 


14 


5.2 


4.8 


6 


3.0 


3.0 


16 


5.5 


5.2 


8 


3.5 


3.5 


18 


5.6 


5.3 


10 


4.0 


3.9 


20 


5.7 


5.4 



GENERAL MATHEMATICS 



At what age do boys grow most rapidly ? At what age do 
girls grow most rapidly ? Is it reasonable to assume that the 
average height of a boy nineteen years old is 5.65 ft.? 

4. The standings of the champion batters from 1900-1907, 
inclusive, are here given for the National and American 
leagues. 

The National League : 
0.384 0.382 0.367 0.355 0.349 0.377 0.339 0.350 



The American League : 
0.387 0.422 0.376 0.355 0.381 



0.329 0.358 0.350 



Graph the data for each league to a convenient scale, both 
on the same sheet. Tell what the lines show. 

5. Draw a temperature chart of a patient, the data for 
which are given below (see Fig. 191). 



Hour . . . 


IJl'.-M. 


7 


8 


9 


10 


11 


12 


1 A.M. 


2 


3 


4 


3 


Temperature 


100.5' 


101 


1015 


103.2" 


102.5 


101.4 


101.3' 


101.3 


101.2 


101.2 


101 


100.7 



6. If possible, get a copy of a temperature curve such as is 
commonly kept in hospitals and explain the graph to the class. 
The class will profit more by your discussion if the curve 
presents the data for a long period. 

7. Be on the lookout for graphic curves which convey infor- 
mation of general interest to your class. In nearly every news- 
paper you will find tables of statistics which may be plotted to 
advantage. Glance occasionally through the "Statistical Ab- 
stract of the United States " or the " Statistical Atlas of the 
United States" (published by the Bureau of Foreign and 
Domestic Commerce), Popular Mechanics Magazine, Popular 
Science Monthly, Scientific American, and so on, with the pur- 
pose of finding interesting graphs. If a lack of time prevents 
class discussion, post these graphs on bulletin boards. 



REPRESENTATION OF STATISTICS 



237 



282. Continuous and discrete series. We may represent 
a continuous change in wealth, in population, in the growth 
of boys and girls, etc. by a smooth curve. Thus, if we read 
four reports of deposits made in a country bank as $20,000 
on January 1, $25,000 on April 1, $18,000 on June 1, 
and $19,000 on September 1, we assume that there was 
a gradual increase of deposits from January 1 to April 1, 
a rather rigorous withdrawal from April 1 to June 1, and a 
slow rally from June 1 to September 1. This is precisely 
the way a physician treats the temperature of a patient, 
even though he may take the temperature but twice per day. 
However, the data of every table cannot be considered as 
continuous between the limits. This fact is clearly illus- 
trated by the following table of Fourth of July accidents : 



YEAR 


KILLED 


INJURED 


TOTAL 


1909 .... ... 


215 


5092 


5307 


1910 


131 


2792 


2923 


1911 


57 


1546 


1603 


1912 


41 


945 


986 


1913 


32 


1163 


1195 


1914 


40 


1506 


1546 











If we were to draw a continuous curve, it would not 
state the facts. Though a few Fourth of July accidents may 
occur on the third or on the fifth of July, we are certain 
that a continuous curve would not represent the facts for 
the rest of the year. Such a collection of items is said to 
be a discrete, or broken, series. A record of wages paid in 
a factory is likely to be a discrete series, for the wages are 
usually (except in piecework) a certain number of dollars 
per week, the fractional parts* being seldom less than 10 <. 
We should find very few men getting odd sums, say, $18.02 
per week. Hence there would be many gaps in the series. 



238 GENERAL MATHEMATICS 

283. Statistics as a science defined. We have now pro- 
gressed far enough for, the student to understand that the 
term " statistics " refers to a large mass of facts, or data, 
which bear upon some human problem. One of the chief 
uses of statistics as a science is to render the meaning of 
masses of figures clear and comprehensible at a glance. 
Statistics gives us a bird's-eye view of a situation involv- 
ing a complex array of numerous instances in such a way 
that we get a picture which centers our attention on a 
few significant relations. Such a view shows how one fac- 
tor in a complicated social or economic problem influences 
another ; in short, it enables us to understand the relation 
between variable (changing) quantities. 

284. The uses of statistics. Statistical studies do not 
exist merely to satisfy idle curiosity. They are necessary 
in the solution of the most weighty social, governmental, and 
economic problems. Do certain social conditions make for 
increase in crime and poverty ? The sociologist determines 
statistically the relations bearing on the question. Are cer- 
tain criminal acts due to heredity ? The biologist presents 
statistical data. Is tuberculosis increasing or decreasing? 
Under what conditions does it increase ? Reliable statistics 
presented by the medical world guide our public hygienic 
policies. Further possibilities of statistical studies in the 
medical world are suggested by the recent work of Dr. Alexis 
Carrel. The work of Dr. Carrel has been widely discussed. 
Though authorities disagree concerning some of the details, 
all will probably agree that the mathematical attack on the 
problem of war surgery is a distinct scientific advance. 

What insurance rates ought we to pay? Statistical 
investigations have determined laws for the expectation of 
life under given conditions which for practical purposes 
are as accurate as the formula for the area of a square. 



REPRESENTATION OF STATISTICS 239 

The business world at times trembles under the threat 
of gigantic strikes that would paralyze all business. Are 
the demands of the men unreasonable ? Are the corpora- 
tions earning undue dividends ? The public does not 
know and will not know until a scientific group of citizens 
present reliable statistics of earnings and expenditures. 

There is now in existence in Washington a nonpar- 
tisan tariff commission which consists of five members ap- 
pointed by the president, which collects statistics and 
makes recommendations to Congress from time to time. 
It is now thought that this commission may tend to do 
away with the. old haphazard methods of handling tariff 
questions. 

How rapidly and with what degree of accuracy should 
a fourth-grade pupil be able to add a certain column of 
figures ? The educator is able to present an answer based 
on tests of more than 100,000 fourth-grade children for 
that particular problem. 

Because of the numerous trained enumerators which 
they employed to cover the world's output, the large spec- 
ulators on the Chicago Board of Trade knew with absolute 
certainty for days in advance of the record-breaking jump 
in the price of wheat in August, 1916. 

We might continue indefinitely to present evidence 
showing that the intelligent reader in any field will profit 
by a knowledge of the elementary principles of statistical 
methods. 

285. Frequency table ; class limits ; class interval. In 
the investigation of a problem it is necessary that the 
data be tabulated in some systematic fashion that will 
enable us to grasp the problem. Suppose we measured 
the length of 220 ears of corn. Let us say the smallest 
ear measures between 5 in. and 6 in., the longest between 



240 



GENERAL MATHEMATICS 



12 in. and 13 in. We could then group the ears by inches, 
throwing them into eight groups, and tabulate the results 
somewhat as follows: 

FREQUENCY TABLE SHOWING LENGTH OF EARS OF CORN 



LENGTH IN INCHES 


NUMBER 
OF EARS 


LENGTH IN INCHES 


NUMBER 
OF EARS 


6-5.99 


2 


9- 9.99 


74 


6-6.99 


4 


10-10.99 ..... 


47 


7-7 99 


20 


11-11.99 ... 


21 


8-8.99 


48 


12-12.99 


4 











70 



CO 



Such an arrangement of data is called a frequency table. 
The ears of corn have been di- niiiimiiiiiiiiiiiimmiiimiimni 
vided into classes. The table 
should be read as follows : 
rt There are two ears measuring 
somewhere between 5 in. and 
6 in., four between 6 in. and 
Tin.," etc. The boundary lines 
are known as class limits, and 
the distance between the two 
limits of any class is designated 40 
as a class interval. The class 
interval in the preceding case 
is 1 in. Class intervals should 
always be equal. 

The facts of the table are 
shown by the graph in Fig. 194. 10 
This graph is the same as the 
bar diagram (Fig. 186) which 
we have drawn, with the excep- 
tion that in this case the bars 
cover the scale intervals. 



so 



20 



7 8 9 10 11 12 13 



FIG. 194. DISTRIBUTION OF FRE- 
QUENCY POLYGON OF 220 EARS 
OF CORN 



REPRESENTATION OF STATISTICS 



241 



EXERCISES 

1. What is your guess as to the shape of the graph if the 
corn were classified into half-inch groups ? into fourth-inch 
groups ? 

2. Are there many very long ears of corn ? many very short 
ears ? How do you tell ? 

3. Into what class interval does the largest group fall ? 

4. The following table of frequency shows the lengths 
of 113 leaves picked from a tree purely at random. (See 
King's Elements of Statistical Method," p. 102.) 



LENGTH OF LEAF IN 
CENTIMETERS 


NUMBER OF 
LEAVES IN 
EACH GROUP 


LENGTH OF LEAF ix 
CENTIMETERS 


NUMBER OF 
LEAVES IN 
EACH GROUP 


3-3.99 .... 


4 


8-8.99 .... 


6 


4-4.99 .... 


5 


9-9.99 .... 


4 


5-5.99 .... 


13 


10-10.99 . . . 


3 


6-6.99 .... 


56 


11-11.99 . . . 


2 


7-7.99 .... 


19 


12-12.99 . . . 


1 



The table should be read as follows : " There are four leaves 
between 3 cm. and 4 cm. long, five between 4 cm. and 5 cm.," etc. 
Graph as in Fig. 194 the data of this table. 

5. The frequency table on page 242 shows the weights of 
1000 twelve-year-old boys, 1000 thirteen-year-old boys, and 
1000 fourteen-year-old boys. (The weights are taken from 
Roberts's "Manual of Anthropometry" and include 9 Ib. of 
clothing in each case. Figures are reduced to the thousand 
basis.) 

6. Study each of the groups for Exs. 4 and 5 as to 

(a) Where the smallest classes are found. 

(b) Where the largest class is located. 

(c) The gradual rise and fall of the figures. 



( i KNERAL M ATHEMATICS 



FREQUENCY TABLE OF WEIGHTS SHOWING 1000 TWELVE- 
YEAR-OLD BOYS, 1000 THIRTEEN-YEAR-OLD BOYS, AND 1000 
FOURTEEN-YEAR-OLD BOYS 



WEIGHTS IN 
POUNDS 


NUMBER OF 

12-YEAR-OLD BOYS 


NUMBER OF 
13-YEAR-OLD BOYS 


NUMBER OF 
14-YEAR-OLD BOYS 


49-56 


4 








56-63 


24 


6 





63-70 


118 


38 


'3 


70-77 


233 


100 


38 


77-84 


273 


225 


41 


84-91 


221 


256 


130 


91-98 


79 


187 


228 


98-105 


36 


112 


^47 


105-112 


12 


43 


118 


112-119 





17 


82 


119-126 





16 


29 


126-133 








12 


133-140 








18 



NOTE. The fact that the third column in the table lacks 54 boys of 
totaling 1000 is because 54 boys in this group weighed over 140 pounds. 
The 49-56 means from 49 up to but not including 56'. Wherever the 
last number of a class is the same as the first number of the next class, 
the first class includes up to that point, but does not include that point. 

Construct the graph (similar to Fig. 194) for each of the 
groups of the table above. 

7. The following test on the ability to use the four funda- 
mental laws in solving simple equations was given to 115 first- 
year high-school students, the time given for the test being 
fifteen minutes. 

DIRECTIONS TO PUPIL 

Find the value of the unknown numbers in each of the following 
equations. Do not check your results. Work the problems in order 
if possible. If you find one too difficult, do not waste too much time 
on it, but pass on to the next. Be sure that it is too difficult, however, 
before you pass on. Do not omit any problem which you can solve. 



243 



1. x + 3 = 7. 

2. 2 #=4. 

3. 2 fc + 7 = 17. 

4. z - 2 = 3. 

5. 2 x - 4 = 6. 

. = , 

7 5 y _ 15 



THE TEST 

13. 16 y + 2 y - 18 y + 2 ij = 22. 

14. 7x + 2 = 3z + 10. 



. 
o 

10. 3.c + 4| =9. 

17. 5.3 y + 0.34 = 2.99. 

18. 0.5x-3 = 1.5. 

19. 3 x -9^ = 17.5 

20. 7 y + 20 - 3 y = 60 + 4 y + 40 - 8 //. 

21. 



66 18 3 






11. + 1 = 6. 
4 

12. |-4=10. 



24. 



25 a; 



The results of the test are given by the following table 
of frequency. The student should study it carefully. 



NUMBER OF 

EXAMPLES 


NUMBER OF 
STUDENTS 
ATTEMPTING 


NUMBER OF 
STUDENTS 
SUCCESSFUL 


NUMBER OF 
EXAMPLES 


NUMBER OF 
STUDENTS 
ATTEMPTING 


NUMBER OF 
STUDENTS 
SUCCESSFUL 











13 


1 


11 


1 





1 


14 


4 


15 


2 





1 


15- 


4 


8 


3 








16 


10 


10 


4 








17 


13 


3 


5 








18 


13 


7 


6 





3 


19 


8 


1 


7 





4 


20 


22 


2 


8 


1 


5 


21 


13 





9 





9 


22 


18 





10 





14 


23 


6 


1 


11 





13 


24 


1 





12 


1 


7 









244 GENERAL MATHEMATICS 

Explanation. The table consists of two parts, of which the first 
part is the first, second, fourth, and fifth columns, which should be 
read, " Of the one hundred and fifteen students taking the test one 
tried but 8 examples, one attempted 12, one attempted 13, four 
attempted 14," etc. The second part consists of the first, third, 
fourth, and sixth columns and should be read, " Of the one hundred 
and fifteen students taking the test one solved only 1 problem cor- 
rectly, one solved only 2 correctly, three solved 6 correctly, four 
solved 7 correctly," etc. 

8. Construct a graph showing the facts of the table given 
on page 243 for the number of students attempting. 

9. Under the directions of your instructor take the test 
in Ex. 7. 

10. Ask your instructor to give you a frequency table show- 
ing the number of attempts and successes in the test taken by 
your class and determine how the work done by your class 
compares in speed and accuracy with that done by the one 
hundred and fifteen students in the test described in Ex. 7. 

286. Measure of central tendencies ; the arithmetic 
average. A frequency table and a frequency graph help 
us to understand a mass of facts because they show us the 
distribution of the items, so that we see where the largest 
groups and the smallest groups fall. The graph shows us 
the general trend of the facts. The large groups assume 
importance. We need terms to describe the central tend- 
ency. Often the word " average " is used to meet this 
need. Such a term is helpful in making a mass of facts 
clear. Thus, a group of farmers could not possibly learn 
much about a field of corn if we read a list to them show- 
ing the length of every ear in a field. But they would 
get some idea of what yield to expect if told that the 
average length of an ear is 91 in. They could certainly 
give a fair estimate of the yield if in addition we told 



REPRESENTATION OF STATISTICS 245 

them that on tha average a row in a square 40-acre field 
grew 620 stalks. We shall presently learn that the word 
" average," as commonly used, is not correct. The phrase 
" arithmetic average " means the quotient obtained by divid- 
ing the sum of all the items by the number of items. Thus, 
to find the average mark obtained by your class on a test 
we need to add the marks of all the students and divide 
by the number of students in the class. If two or more 
students obtain the same mark (say 70), we can shorten 
the first step of the process by multiplying the mark by the 
number of times it occurs instead of adding 70 five times. 
This means that in a frequency table a student must re- 
member to multiply each item by its frequency before adding. 
When the size of the item is only approximately known, 
the mid-point of the class interval is taken to represent 
the size of each item therein. To illustrate, suppose that 
we should try to find the average number of problems 
attempted in the simple-equation test. We shall suppose 
that three students report that they attempted 6 problems. 
This does not really mean that all three exactly completed 
6 problems when time was called. In all probability one 
had made a slight start on number 7, the second was about 
in the middle of number 7, and the third had almost 
completed 7. Of course the number of students is too 
small to make this certain, but if we should take a larger 
number of students (say thirteen), in air probability there 
would be as many who were more than half through with 
the seventh problem as there would be students less than 
half through. Hence, to find the average we say that the 
thirteen students attempted 6i and not 6, as they reported. 
To illustrate: 

Find the average number of equations attempted by a class 
on the simple-equation test if two students report 5 problems 



24G GENERAL MATHEMATK S 

attempted, four report 6, five report 7, three report 8, and two 
report 9. 

Solution. 2x5 = 11 

4 x 6 = 26 

5 x 7 1 = 37.:. 
:5 x 8i = 2').-, 
2x9* = 19 

16 119 

110 -=-16 =7.4. 

Therefore the average number of problems attempted by the 
class is 7.4. 

The point is that the series of facts in the table is not 
a discrete series, as one would at first be inclined to think, 
but a continuous series. 

EXERCISES 

*1. Calculate the average number of equations attempted by 
your class in the simple-equation test (Art. 285). 

2. Using the table of Ex. 5, Art. 285, find the average 
weight of the twelve-year-old boys ; of the thirteen-year-old 
boys ; of the fourteen-year-old boys. 

3. Find the average length of the 113 leaves in Ex. 4, 
Art. 285. 

4. Find the average length of the 220 ears of corn of the 
first table in Art. 285. 

5. Find the average number of Fourth of July accidents 
for the six years of the table of Art. 282. 

6. Compare the averages of the champion batters for the 
last eight years (Ex. 4, Art. 281). 

7. At Minneapolis the 7 A.M. temperature readings for 
the ten days beginning February 1 were as follows : 5, 3, 
5, 3, 7, 9, 8, 2,0, 6. Find the average 7 A.M. 
temperature reading for the period. 



REPRESENTATION OF STATISTICS 247 

8. Find the average of the following temperatures: 8 A.M., 
-6; 9 A.M., -5; 10A.M., -2; 11 A.M., -1; 12 M., 
-2; IP.M., -4; 2 P.M., -6; 3 P.M., -7; 4 P.M., -7; 
5 P. M., - 5 ; 6 P. M., - 2 ; 7 P. M., - 1. 

9. Find the average church contributions according to the 
following frequency table. 

TABLE OF CHURCH CONTRIBUTIONS 



INDIVIDUAL 
CONTRIBUTIONS 


NTMBER OF 
CONTRIBUTORS 


INDIVIDUAL 
CONTRIBUTIONS 


NUMBER OF 
CONTRIBUTORS 


No contribution . 
1 cent 


2 

23 


10 cents . . . 
25 cents 


13 

9 


5 cents 


42 


$50 


1 











287. Disadvantages of the arithmetic average. Some of 
the preceding exercises suggest that there are certain 
objections to the arithmetic average. For example, it 
means little to say that the average church contribution 
in Ex. 9, Art. 286, is 62 cents. People ordinarily use 
the word " average " thinking it means the most usual 
occurrence ; that is, the common thing. As a matter of fact 
nobody gave 62 cents, and only one person gave as much 
as that. The objection to the arithmetic average is that it 
gives too much emphasis to the extreme items. To illustrate 
more fully: A boy who has just finished an elementary 
surveying course learns that the average weekly wage of a 
railway-surveying group is $23. This is very encduraging 
until an analysis shows him that the chief engineer gets $55 
a week ; his assistant, |30 ; and all others but $15. To say 
that the average weekly earning of ten men working in an 
insurance office is $30 a week may be misleading, for one 
man may be a $5000-a-year man, in which case the usual 
salary is much lower than $30 per week. Other objections 



248 GENERAL MATHEMATICS 

to the arithmetic average are the following: (a) it cannot 
be located either in a frequency table or in a frequency 
graph ; (b) it cannot be accurately determined when the 
extreme items are missing; (c) it is likely to fall where 
no item exists (for example, a sociologist may discover 
that the average-size family in a given community has 4.39 
members, though such a number is evidently impossible). 
For these reasons it is desirable to have some other 
measure of tht central tendency of a group. 

288. Central tendency ; the mode. One of the most use- 
ful measures is the mode. It may be denned as the scale 
interval that has the most frequent item, or we may say it 
is the place where the longest bar of a bar diagram is 
drawn. The term describes the most usual occurrence, or 
the common thing. The popular use of the term " average " 
approximates the meaning of the word. When we hear of 
the average high-school boy he is supposed to represent 
a type one who receives exactly the most common mark 
of his classes, is of the most common athletic ability, 
spends the most common amount of time in study, shows 
the most common amount of school spirit, wastes the most 
common amount of time, is of the most common age, etc. 
It is obvious that no such high-school boy can be found. 
Though a boy may possess some of these attributes he is 
certain to differ from the common type in others. 

The word " average " is thus incorrectly used for " mode," 
which means the common type. Thus the mode in the 
church-contribution table (Ex. 9, Art. 286) is five cents. 
More people in this church gave a nickel than any other 
coin. The mode in the frequency 'table for the simple- 
equation test for attempts (Ex. 7, Art. 285) is twenty 
examples. In the test more students were at this point 
when time was called than at any other point. 



REPRESENTATION OF STATISTICS 249 

EXERCISES 

1. Find the mode for your class in the frequency table for 
successes in the simple-equation test of Ex. 7, Art. 285. 

2. From the table of Ex. 5, Art. 285, find the mode for the 
weight of the twelve-year-old boys ; of the thirteen-year-old 
boys ; of the fourteen-year-old boys. 

NOTE. The student is expected merely to glance at the tables 
to see at what scale interval the greatest number of items are found. 

3. From the table of Ex. 4, Art. 285, find the mode for the 
113 leaves. 

4. From the first table of Art. 285 find the mode for the 
220 ears of corn. 

5. From the table in Art. 282 find the mode for the Fourth 
of July accidents. 

289. Advantages and disadvantages of the mode. The 

advantages of the mode may be summarized as follows: 

(a) the mode eliminates the extremes. In the results 
of an examination the mode is not affected by the occa- 
sional hundred or zero marks. The salary of the super- 
intendent of a division of a shop does not affect the mode ; 

(b) to the ordinary mind the mode means more than 
does an average. It means more to say that the modal size 
of classes in a high school is 15 than to say that the average 
size is 17.24, first, because there is not a single class that 
actually has the latter number ; and, second, because a few 
large freshman classes in city high schools may tend to 
increase considerably the average number. In making laws 
we shall do the greatest good to the greatest number if we 
keep the mode in mind and not the average. Unfortunately 
street cars are built for the average number carried, 



250 GENERAL MATHEMATICS 

not for the modal number ; hence the " strap hanger." 
The manufacturer of ready-made clothing fits the modal 
man, not the average man. The spirit of a community's 
charity fund is far more evident in the mode than in the 
average. 

A disadvantage of the mode is that there are a large 
number of frequency tables to which it cannot be easily 
applied. In such cases we have an irregular group with no 
particular type standing out, and the mode is difficult to 
find, as will be illustrated presently. 

290. Central tendency ; the median. If a number of ob- 
jects are measured with reference to some trait, or attribute, 
and then ranked accordingly, they are said to be arrayed. 
Suppose that your instructor gives an examination which 
really tests mathematical ability, and that after the results 
are announced the students stand in line, taking the 
position corresponding to their marks on the examination ; 
that is, the student with least mathematical ability at the 
foot of the class, the one next in ability next to the foot, 
etc. The class is then arrayed. If any group of objects is 
arrayed, the middle one is known as the median item. If 
your class had twenty-three pupils standing in the order 
of their ability, the twelfth pupil from the foot or the head 
of the class is the student whose mark is the median mark. 
There are just as many below as above him in ability. The 
median is another measure of the central tendency of a 
group. If there is an even number of items, the median 
is said to exist halfway between the two middle items. 
Thus, if your class had twenty-two pupils, the mark half- 
way between that of the eleventh and that of the twelfth 
student from either end would be called, the median mark. 
The meaning is further illustrated by the exercises given 
on pages 251 and 252. 



KEPKESENTATION OF STATISTICS 



251 



1. Find the median wage in the following table of the 
weekly wage of the workers in a retail millinery shop. 1 



WEEKLY WAGE 


NCMBER OF 
WORKERS 


. 
WEEKLY WAGE 


NUMBER OF 
WORKERS 


$4-5 . 


g 


$11-12 





o-O 


18 


12-13 


5 


(5-7 


16 


13-14 . . ' . 





7-8 


12 


14-lo 


6 


8-9 


7 


15-16 


18 


9-10 


8 


16-17 .... 


5 


1(H11 


5 















The table above shows the wages of one hundred girls. 
We are asked to find a weekly wage so that we shall be 
able to say that one half the girls in this shop receive less 
than this sum and one half receive more than this sum; 
that is, we want a measure of the group. 

In the first place, the student should notice that the 
arithmetic average $10.05 seems too high to be represent- 
ative, for there are too many girls working for smaller 
sums. In the second place, the mode is unsatisfactory. 
The wage $15 to $16 seems to be a mode, but there are 
more girls working for about $5, $6, or $7 ; hence neither 
the arithmetic average nor the mode has very much mean- 
ing, so we proceed to locate the median. 

There are one hundred girls in the shop ; hence we must 
find a wage halfway between that of the fiftieth and that 
of the fifty-first girl from the lowest wage. Adding the 
number of the first four groups of girls (3 + 15 + 16+12) 
gives us forty-six girls and takes us to the 8-dollar wage. 
We need to count four more of the next seven, who are 

1 For actual facts see "Dressmaking and Millinery," in "The Cleve- 
land Survey," page 63. The table was adapted to meet the purposes of 
the exercise. 




252 GENERAL MATHEMATICS 

getting between #8 and $9. The table assumes that the 
series is continuous (piecework ?) ; hence the next seven 
are distributed at equal distances between $8 and $9. We 
may think of the seven girls as being distributed graphi- 
cally, as shown in Fig. 195. 

The graph makes clear the assumption that the first girl 
(the forty -seventh) earns a sum which is between 8 and 
$8| ; we assume that the wage is at the mid-point of this in- 
terval, or $8-Jj. Similarly, the second girl (the forty-eighth) 
earns a sum between $8| and 
$8f and we assume the wage 
to be at the mid-point of this 
interval, or |8^. In like man- 
ner the wage of the forty- F 195 

ninth girl is $8 T 5 , the fiftieth 

girl |8 T 7 j, and the fifty-first girl $8 T 9 ? . Midway between 
the mid-points of the fiftieth and the fifty-first wage is 
halfway between $8 T 7 and $8-^, or $8|. Hence the 
median is $8 plus $|, or $84, for this wage is halfway 
between the wage of the fiftieth girl and that of the fifty- 
first. The student should study this graph until this 
point is clear. He should note that the average is found 
by calculating, the mode by inspection, and the median by 
counting. Merely count' along the imagined scale until a 
point is found that divides the item into two equal groups. 
Since a wage problem usually involves a discrete series 
(why?), a more practical illustration of the principle is 
given below. 

2. Find the median for the attempts of the one hundred and 
fifteen students in the simple-equation test in Ex. 7, Art. 285. 

Solution. We must find the number of equations attempted by 
the fifty-eighth student from either end, for he will be the middle 
student of the one hundred and fifteen. Counting from the top of 



REPRESENTATION OF STATISTICS 253 

the table (p. 243), we get fifty-five pupils who hive finished 19 
equations. We need to count 3 more to get the fifty-eighth pupil. 
There are twenty-two more who were somewhere in the twentieth 
equation when time was called. If we assume, as we did in finding 
modes, that the twenty-two students are at equal spaces through- 
out the twentieth equation, then the median is 19 equations plus 
? \ of an equation, or just over 19.1 equations. 

EXERCISES 

1. State the rule for finding the median, as developed in 
the two preceding exercises. 

2. In Ex. 5, Art. 285, find the median weight for 1000 
twelve-year-old boys; 1000 thirteen-year-eld boys; 1000 
fourteen-year-old boys. 

3. In Ex. 4, Art. 285, find the median leaf and its measure 
in the array of 113 leaves. 

4. Find the median for the 220 ears of corn (Art. 285). 

291. Limitations of statistics. There is a common saying 
among nonscientific people that anything can be proved by 
means of statistics. Experience lends conviction to the 
homely saying " Figures' do not lie, but liars will figure." 
This belief is due to the fact that figures have deceived the 
public either by being dishonestly manipulated or by being 
handled unscientifically. A table dishonestly manipulated 
or based on unreliable data appears at first glance as con- 
vincing as the work of a trained scientist. The public does 
not find it possible to submit every piece of evidence to a 
critical study and resents such deceptions as those referred 
to above. 

As a beginning the student should determine (1) the 
reliability, and training of those 'who gathered the facts; 
(2) how and when they were gathered ; (3) to what extent 



254 GENERAL MATHEMATICS 

the statistical studies have been exposed to the critical 
judgment of trained experts; (4) to what extent similar 
studies show similar results. 

292. Law of statistical regularity. In calculating the 
value of the farm lands in Indiana it is by no means 
necessary to evaluate and tabulate every acre in the state. 
To find out the average size of a twenty-five-year-old man 
in New York it is not necessary .to measure and tabulate 
every man in the city. The " Statistical Abstract of the 
United States" (published by the Bureau of Foreign and 
Domestic Commerce) states the value in dollars of hogs, 
sheep, and cattle- produced in 1918, but this does not mean 
that this total is obtained by tabulating every individual 
animal. To find out how fast on an average a twelve- 
year-old Chicago boy can run 100 yards we would not need 
to hold a stop watch on every boy. In fact, a few chosen 
in each class in each school building would probably give 
us an average that would be identical with an average 
obtained from the whole group. This is due to a mathe- 
matical law of nature which states that if a reasonable num- 
ber of individual cases are chosen " at random " from among 
a very large group, they are almost sure, on the average, to pos- 
sess the same characteristics as the larger groups. The phrases 
" at random" and "reasonable number" make the law appear 
somewhat vague. King, in "Elements of Statistical Method," 
illustrates the law as follows : "If two persons, blindfolded, 
were to pick, here and there, three hundred walnuts from 
a bin containing a million nute, the average weight of the 
nuts picked out by each person would be almost iden- 
tical, even though the nuts varied considerably in size." 
Gamblers use the principle just illustrated when they have 
determined how many times a given event happens out of 
a given number of possibilities. They are thus able to ply 



REPRESENTATION OF STATISTICS '255 

their craft continuously and profitably on a small margin 
in their favor. This principle is the basis of all insur- 
ance ; thus it is possible to predict with a great degree 
of accuracy how many men of a given age out of a given 
one thousand will, under ordinary conditions, die during 
the next year. The law of statistical regularity is very 
extensively employed in the Census Bureau. The totals 
are usually estimates based on careful study of sufficient 
representative cases. 

However, the student should be critical of the phrase 
'" at random." It is not asserted that any small group 
will give the same results as a measurement of the whole 
group. Thus, if we measured the height of the first 
four hundred men that passed us as we stood at the 
corner of Randolph and State Streets, Chicago, we could 
not be sure of getting an average that would accurately 
represent the city. Any number of events might vitiate the 
results ; for example, the Minnesota football team might 
be passing by, or a group of unusually small men might 
be returning from some political or social meeting limited 
to one nationality. The sampling should be represent- 
ative ; that is, sufficiently large and at random (here and 
there). The larger the number of items, the greater the 
chances of getting a fair sample of the larger group of 
objects studied. 

*293. The law of inertia of large numbers. This law 
follows from the law of statistical regularity. It asserts 
that when a part of a large group differs so as to show a 
tendency in one direction, the probability is that an equal 
part of the same group has a tendency to vary in the opposite 
direction; hence the total change is slight. 

294. Compensating and cumulative errors. The preced- 
ing laws are also involved in a discussion of errors. If 



256 GENERAL MATHEMATICS 

the pupils in your school were to measure carefully the 
length of your instructor's desk, the chances would be that 
as many would give results too large as too short. 

The estimates of a thousand observers of crop conditions 
which are summarized or graphed in a volume such as 
the " Statistical Atlas " (published by the Department of 
Commerce) tend to approximate actual conditions. These 
are illustrations of compensating errors. " In the long run 
they tend to make the result lower as much as higher." 
This type of error need not concern us greatly, provided 
we have a sufficient number of cases. 

However, we need to be on our guard against a con- 
stant or cumulative error. If we use a meter stick that is 
too short, we cannot eliminate the error by measuring a 
very long line. A watch too fast could not eventually be 
a correct guide. A wholesaler who lost a little on each 
article sold could not possibly square accounts by selling 
large quantities. 

The value of a mass of facts involving a constant error 
is seriously vitiated. Hence the student should be con- 
stantly critical in his effort to detect this type. 

EXERCISE 

*Draw as accurately as possible on the blackboard a line 
segment a certain number of inches in length. Ask as many 
as from forty to fifty schoolmates, if possible, to stand on a cer- 
tain spot and estimate the length of the line. Find how many 
estimated the line too long ; how many estimated it too short. 

HINT. The work must be done carefully. Have each student 
estimate four times ; that is, estimate, look away, estimate, etc. Reject 
estimates of all students who do not comply seriously with your 
request. How many estimated the line too long ? How many esti- 
mated it too short ? Report the results to your class. 



KEPRESE^sTATION OF STATISTICS 



257 



*295. Normal distribution. The student has observed 
the regular rise and fall of the numbers of the frequency 
tables and the regular rise and fall of the graphs which 
express these relations. It appears that the large majority 
of the items are usually grouped, and as the distance from 
this point of grouping becomes greater, the items become 
rapidly fewer in number. If we measure ability to solve a 
set of 24 simple equations, we shall discover that the large 
majority of any representative class can attempt a little 
more than 19 in 15 minutes. Only a few, possibly none, 
will try all 24 problems; and only a few, probably none, 
will have attempted 



less than 6. 
measure any 



If we 

single 




FIG. 196. FREQUENCY CHART OF SCALLOP 

SHELLS PILED ACCORDING TO THE NUMBER 

OF RIBS. (AFTER BRINTON) 



human trait, we shall 
discover the same 
tendency in the graph 
of the results toward 
a bell-shaped curve. 
Whether we measure ability to spell, ability to add a column 
of figures, ability to throw a baseball, the distance boys can 
broad-jump, always there are a few very good at it and 
a few very poor at it, with the tendency of the great 
majority to possess but mediocre ability in a particular trait. 
Nature shows the same tendency. The length of leaves, 
the height of cornstalks, the length of ears of corn, etc. 
give curves similar to the preceding. Note the piles of 
scallop shells in Fig. 196. The shells are sorted into 
separate piles according to the number of ribs. The piles 
(from left to right) have respectively 15, 16, 1.7, 18, 19, 
and 20 ribs. How do you think these piles would have 
looked if a greater number of shells (say several hundred) 
had been used? 



258 



GENERAL MATHEMATICS 



The same tendency is observed in economic affairs. 
Thus, if we measured the income of the ordinary agri- 
cultural community, we should find out of a thousand 
persons only a few whose income is less than $300 per year, 
only a few, if any, with an income over $2500, and the 
rest grouped and 
tapering between 
these limits. 

When the rise 
and fall is regu- 
lar (that is, the 
curve falls regu- 
larly on both sides 
from the mode), 
the distribution is 
likely to approxi- 
mate what we call 
a normal distribu- 
tion, and the curve 
is called a normal 
distribution curve. 

A normal dis- 
tribution is illus- 
trated by the 
table and diagram 
(Fig. 197) given here, which represents the heights from 
actual measurement of four hundred and thirty Eng- 
lish public-school boys from eleven to twelve years 
of age. 1 

It will be seen that the numbers conform to a very 
uniform rule: the most numerous groups are in the middle, 
at 53 in. and 54 in., while the groups at 51 in. and 56 in. 

1 From Roberts's "Manual of Anthropometry," p. 18. 




FIG. 197. 



PHYSICAL PHENOMENA ILLUSTRATE 
NORMAL DISTRIBUTION 



REPRESENTATION OF STATISTICS 



259 



are less in number, those at 50 in. and 57 in. are still fewer, 
and so on until the extremely small numbers of the very 
short and very tall boys of 47 in. and 60 in. are reached. 
It is shown that the modal, or typical, boy of the class 
and age given is 53.5 in., and since he represents the most 
numerous group, he forms the standard. 

The curve would probably be smoother if more boys 
were measured or grouped into half -inch groups. As it is, 
it approximates very nearly a normal distribution. 

Of course it is not asserted that every distribution is 
of this type. There is merely a tendency in chance and 
in nature to produce it. There are many causes which 
make distribution irregular, as we shall presently see. 

*296. Symmetry of a curve. The graph in Fig. 198 
shows the height of 25,878 American adult men in inches. 
This curve, like the one 
of Art. 295, is more reg- 
ular than most curves 
which we have studied. 
It would probably be 
much smoother if the 
class interval were one- 
fourth inch. If we draw 
a perpendicular AK from 
the highest point of the curve, we may think of this as an 
axis around which the rectangles are built. The curve 
to the right of this axis looks very much like the part to 
the left. In this respect we say the curve is almost 
symmetrical. 

Symmetry of figures may be illustrated by the human 
head, which is symmetrical with respect to a plane midway 
between the eyes and perpendicular to the face ; thus the 
left eye and the left ear have corresponding parts to the 




FIG. 198. HEIGHT OF MEN. (AFTER 

THORNDIKE, " MENTAL AND SOCIAL 

MEASUREMENTS," p. 98) 



260 



GENERAL MATHEMATICS 



right of this ti.ris of symmetry. Note that the parts are 
arrayed in reverse order. 

Other familiar illustrations of symmetry are (1) the 
hand and the image obtained by holding the hand in front 
of a plane mirror ; (2) words written in ink and the im- 
print of those words on the blotting paper with which they 
are blotted : (3) our clothes, which are largely built on the 
principle of .symmetry ; (4) the normal distribution curve. 

In architecture, in art, and in higher mathematics the 
principle of symmetry is very important. 

*297. Skewness of a curve. The term "skewness" de- 
notes the opposite of symmetry and means that the items 
are not symmetrically distributed. The curve is not of the 
bell-shaped form. It is higher either above or below the 
mode than a sense of symmetry would have us expect. 
To illustrate: Snppose that the incomes of all the people 
living in a certain community were tabulated as follows : 



INCO.MK ix DOLLARS 


XfMBER OF 

PERSONS 


INCOME ix DOLLARS 


N I'M HER OF 
PERSONA 


0-500 . . . 


20 


3500-4000 . . . 


4 


500-1000 . . . 


36 


4000-4500 . . . 


3 


1000-1500 . . . 


20 


4500-5000 . . . 


o 


1500-2000 . . . 


12 


5000-5500 . . . 


1 


2000-2500 . . . 


6 


5500-6000 . . . 


1 


2.300-3000 ... 


5 


6000-6500 ... 


1 


3000-3500 . . . 


4 





The graph (Fig. 199) of this table is not symmetri- 
cal, but is skewed toward the lower side. The meaning of 
skewness is clearly shown by the graph. The graph no 
longer presents the normal, symmetrical, bell-shaped form ; 
the base is drawn out to a greater extent on the one side 
than on the other. 



REPRESENTATION OF STATISTICS 



261 



Distribution is often affected by laws which are dis- 
turbing factors in the situation. Thus, in investigating 
the wages of carpenters we should expect a few to get 
high wages, say 90^ per hour, and a few very low, say 40 < 
per hour, and the rest to be grouped, according to ability, 
between these limits. However, by agreement between 
unions and contractors, carpenters' wages are fixed in most 




FIG. 190. GRAPH SHOWING SKE\VNESS OF A CURVE 

cities at a price somewhere between 60^ and 85<. Hence 
we should have but one interval in a distribution table, 
for a particular city, say Minneapolis, showing that all 
carpenters get 75 < per hour. 

R 

298. The graph of constant cost relations. 1 Graphs may be 
constructed and used as " ready reckoners " for determining 

1 Teachers may find it desirable to take up the graphing of formulas 
from science at this point; for example, the graph of the centigrade- 
Fahrenheit formula. However, the authors prefer to use a simpler 
introductory exercise here for the sake of method and to take up 
more purposeful formulae in the next chapter. 



262 



GENERAL MATHEMATICS 



costs of different quantities of goods without computation. 
This is shown by the following example: 

If oranges sell at 30 $ a dozen, the relation between the num- 
ber of dozens and the cost may be expressed by the equation 
c = 30 c/, where d is the number of dozens and c the cost 
per dozen. If values are given to d, corresponding values 
may be found for c, as given in the following table : 



d 





1 


2 


3 


4 


5 


10 


11 


c 





30 


(iO 


90 


120 


150 


300 


330 



On squared paper draw two axes, OX and OF, at right 
angles. On OY let a small unit represent 1 doz., and on 
OX let a small unit represent 10 $. Then, on the 30 < line 



10 



50 100 150 200 250 300 

FIG. 200. THE GRAPH OF A COST FORMULA 



X 



mark a point representing 1 doz. On the 20$ line mark 
a point representing 3 doz. Draw a line through the points 
thus marked. It is seen that this line, or graph (Fig. 200), 
is a straight line. 

By looking at this price curve we can get the cost of 
any number of dozens, even of a fractional number. For 
example, to find the cost of 6 doz. observe the point where 



263 

the horizontal line six small units up meets the price 
curve ; observe the point directly beneath this on the 
axis OX', this is eighteen small units from and hence 
represents $1.80. Similarly, the cost of 8 doz. is seen 
to be $2.55. 

EXERCISES 

1. By means of the graph in Fig. 200 determine the cost 
of the following: 9 doz.; 11 doz.; 2^ doz.; 3^ doz.; 10|doz.; 
5| doz. ; 3^ doz. 

2. If eggs sell at 45$ a dozen, draw the price graph. 

3. On the price graph drawn for Ex. 2 find the cost of 4 doz. ; 
3 doz.; 10 doz. ; 3^doz.; 5^ doz. ; 4^ doz. 

4. Draw a price graph for sugar costing 10| $ a pound. 

5. On the graph drawn for Ex. 4 find the cost of 11 lb.; 
31 lb.; 6|lb.; 10 lb. 

6. Construct a graph which may be used in calculating the 
price of potatoes at $2.10 per bushel. 

7. Use the graph of Ex. 6 to find the cost of 3 bu.; 4|- bu.; 
2 bu. 3 pk. ; 5 bu. ; 5 bu. 3 pk. 

8. Since the graphs in Exs. 1~7 are straight lines, how 
many of the points would have to be located in each case in 
order to draw the line ? Should these be taken close together 
or far apart, in order to get the position of the price graph 
more nearly accurate ? Why ? 

299. Graphs of linear equations ; locus ; coordinates. As 
shown in the preceding sections the relation between two 
quantities may be expressed in three ways: (1) by an 
ordinary English sentence, (2) by an equation, or (3) by 
a graph. The graph is said to be the graph of the equation. 
A graph may be constructed for each equation that we 



264 



GENERAL MATHEMATICS 



have studied to date. The process of drawing the graph 
of an equation will be given in this article. 

Let the equation be y = 2 x + 3, which we shall suppose 
is the translation of some sentence which states some 
definite practical rule; for example, the cost of sending a 
package by parcel post into a certain zone equals two cents 
per ounce plus three cents. We want to draw a graph 
for the equation y = 2 x + 3. 



EXERCISES 

1. What is the value of y in the equation y = 2x + 3 when 
x equals ? when x equals 1 ? when x equals 2 ? when x equals 
3 ? when x equals 2 ? when x equals 3 ? 

2. Fill in the following table of values of x and y for the 
equation y = 2 x + 3. 



X 





1 


2 


3 


4 


5 


6 


7 


- 1 


- 2 


- 3 


- 4 


y 


3 


5 


7 






















We are now ready to transfer the data of Ex. 2 to 
squared paper. The process does not differ very much 
from our work in frequency tables except that usually in 
graphing equations we need to consider both positive and 
negative numbers. For the sake of method we shall extend 
the discussion to cover this point. Two axes, AA'' and 
)')"' (Fig. 201), are drawn at right angles^and meet 
at 0. Corresponding to each set of values of x and y a 
point is located, the values of x being measured along or 
parallel to XX', and the values of y along or parallel to 
YY'. -Positive values of x are measured to the right of 
YY' and negative values to the left ; positive values of y 
are measured above XX' and negative values below A'A"'. 



REPRESENTATION OF STATISTICS 



265 



m 



I 



For example, the point A corresponding to x ^= 1 and y = 5 
is obtained by measuring one space to the right and five 
spaces upward. The point B corresponding to x = 1 and 
y = 1 is obtained by measuring one space to the left and 
one space upward. The 
point C corresponding 
to x = 3 and y = 3 
is obtained by measur- 
ing three spaces to the 
left and three spaces 
downward. The x and 
y values of each point 
are called the coordi- 
nates of that point. 

Continue finding 
points which represent 
corresponding parts of 
numbers in the table. 
tt soon becomes ap- 
parent that all the points seem to lie on a straight line. 
Hence we need to plot only two points to be able to 
draw the line. However, we are more certain to discover 
possible errors if we plot three points. Why? 

, EXERCISES 

1. Find the values of x and y at the points D and E 
(Fig. 201) by inspection. Determine whether they satisfy 
the equation of the graph. 

2. Select any point in the line and determine whether the 
values of x and y at this point satisfy the equation. 

3. Select any point not on the graph, find the values of x 
and ylat this point, and determine whether they satisfy the 
equation of the graph. 



FIG. 201. GRAPH OF A PARCEL POST 
FORMULA 



260 GENERAL MATHEMATICS 

4. Select any point of the graph and determine whether the 
values of x and y satisfy the equation. 

5. How many points could one find on the line ? 

The preceding exercises illustrate the following facts : 

(a) The coordinates of every point on the line satisfy the 
equation. 

(b) The coordinates of every point not on the line do not 
satisfy the equation. 

These two facts can be proved rigidly in advanced 
mathematics, and they enable us to say that the straight 
line found is the lo<nis (the place) of all points whose coordi- 
nates satisfy the given equation. It is important to observe 
that the idea of a locus involves two things, specified under 
(a) and (b) above. 

Since it appears that the graph of an equation of the 
first degree having two unknowns is a straight line, equa- 
tions of the first degree are called linear equations. 

A line may be extended indefinitely in either direction, 
and there are an indefinitely large (infinite) number of points 
upon a straight line. Since the coordinates of each point on 
the line satisfy the equation of the line, there are an infinite 
number of solutions of a linear equation with two unknowns. 
This fact is evident, also, because for every value of one of 
the unknowns we can find a corresponding value for the 
other unknown. 

ORAL EXERCISES 

1. What is the location (locus) of all points in a plane 
which are at a distance of 5 ft. from a given point P in the 
plane? at a distance of 7% ft. fromP? at a distance of x feet,? 

2. What is the locus of all points in space at a distance 
of 10 ft. from a given point ? 1 cm. from a given point ? 
x yards from a given point ? 



REPRESENTATION OF STATISTICS 



267 



3. What is the locus of all points in a plane 3 in. distant from a 
given straight line in the plane ? 5| in. distant ? y inches distant ? 

4. What is the locus of all points in space at a distance of 
4 in. from a given straight line ? 6 cm. ? y feet ? 

5. What is the locus of all points in a plane equally dis- 
tant from two given parallel lines in the plane ? 

*6. What is the locus of all points in space equally distant 
from two given parallel lines ? 

7. What is the locus of all points in a plane 6 in. distant from 
each of two given points in the plane which are 10 in. apart ? 

8. What is the locus of all points 3 ft. from the ceiling of 
your classroom ? 

*9. What is the locus of all points in a plane 5 in. distant 
from a line segment 7 in. long in the plane ? 

*10. What is the locus of all points in space 5 in. distant 
from a line segment 10 ft. long ? 

300. Terms used in graphing a linear equation. Certain 
terms used in mathematics ,in connection with graphical 
representation will now 
be given and illustrated 
by Fig. 202. The lines 
XX' and YY', drawn at 
right angles, are called 
axes (XX 1 the horizontal 
axis and YY' the verti- 
cal axis). The point is 
called the origin. From P, 
any point on the squared 



-x 



2 






paper, perpendiculars are 

drawn to the axes: the FlG - 202 ' I""MRATISG K TERMS 

USED IN PLOTTING A POINT 

distance PM is called the 

ordinate of P, and the distance PN is called the abscissa of P ; 



2b'8 GENERAL MATHEMATICS 

together they are called the coordinates of P. The axes are 
called coordinate axes. The scale used is indicated on the 
axes. The distances on OX and Y are positive ; those on 
OX' and on OY' negative. The abscissa of P is 2 and the 
ordinate is 2*-; the point P is called the point (%, 2). 
Notice that the abscissa is written first and the ordinate 
second. Finding a point on a graphic sheet which corre- 
sponds to a given pair of coordinates is called plotting 
the point. 

EXERCISES 

1. What is the abscissa of point A ? B ? C ? D ? E ? (Fig. 202). 

2. What is the ordinate of the point A? B? C? D? E :' 
(Fig. 202.) 

3. Give the coordinates of points A, B, C, D, E (Fig. 202). 

4. On a sheet of graphic paper draw a set of coordinate axes 
intersecting near the center of the paper, and plot the following 
points : (2, 4), (5, 2), (4, - 2), (- 3, 4), (- 3, - 2),(- 2J, 3J). 

5 . Compare the process of plotting points with the numbering 
of houses in a city. 

6. On a sheet of graphic paper locate the points A (2, 2), 
B(5, 3), C(2, 7), and D(5,.8). What kind of figure do you 
think is formed when the points A, B, C, and 7)are connected? 
Draw the diagonals of the figure, and find the coordinates of 
the point where the diagonals intersect. 

301. Summary of the method for the process of graphing 
a linear equation. With Art. 300 in mind we shall now 
illustrate and summarize the process of graphing a linear 
equation. 

Draw the graph of 4 x 3 y = 6. 

(a) Solve the equation for either unknown in terms of the other: thus, 



REPRESENTATION OF STATISTICS 



269 



This throws the equation into a form from which the corre- 
sponding pairs of values are more easily obtained. 



(b) Let 
Then 
And let 
Then 



x 3, etc. 



That is, build a table of corresponding values as follows : (Try to 
get at least two pairs of integral numbers. Why?) 



6 + 3 77 




(c) Plot the points 

corresponding to the 

f t f.j, FIG. 203. GRAPH OP THE LINEAR EQUATION 
pairs oj numbers oj the . _ , 

table (see Fig. 203). 

(d) To check, choose a point on the line drawn and determine 
whether its coordinates satisfy the given equation or plot a third 
pair of numbers in the table. This third point also should fall on 
the- line drawn. 

(e) The two points plotted should not be too near each other. Why f 



270 GENERAL MATHEMATICS 

> 

EXERCISES 

Draw the graphs of the following equations, each on a sep- 
arate sheet of squared paper : 

1. x + y = 7. 5. 5x- 4y = 20. 9. 5x-2y = -3. 

2.2x y = &. 6. 3x + 5y = l5. 10. x + 5 y 12. 

3. 3x- 2y = l2. 7. 5x-2y = W. 11. 2x = 3-4/. 

4. 3x + 2y=6. 8. 6x-4y = 3. 12. 3y = 4 8ar. 

HISTORICAL NOTE. Statistics has attained the dignity of a sci- 
ence during the last fifty years. Its growth goes hand in hand with 
national organization. Even in a crude tribal organization the ruler 
must needs know something of its wealth to determine the taxes or 
tribute which may be levied. Our earliest statistical compilations 
(some time before 3000 B.C.) presented the population and wealth 
of Egypt in order to arrange for the construction of the pyramids. 
Many centuries later (about 1400 B.C.) Rameges II took a census 
of all the lands of Egypt to reapportion his subjects. 

In the Bible we read how Moses numbered the tribes of Israel 
and of the census of the Roman emperor, Augustus Caesar, in the 
year which marked the birth of Christ. 

The Greeks and Romans and the feudal barons of the Middle 
Ages made many enumerations for the purposes of apportioning 
land, levying taxes, classifying the inhabitants, and determining the 
military strength. In all cases except that of the Romans some 
special reason existed for collecting the data. The Romans col- 
lected such data at regular intervals. 

During the Mercantile Age of western Europe the feeling grew 
that it was the function of a government to encourage the measures 
aimed to secure a balance of trade. In order to decide correctly 
concerning the needs of commercial legislation, more detailed infor- 
mation was necessary than had hitherto been gathered. The growth 
in a centralized monarchy further stimulated statistical study. That 
monarch was most successful who could in advance most accurately 
compare his resources with his rivals'. 

In 1575 Philip II of Spain made extensive inquiries from the prel- 
ates concerning their districts. In 1696 Louis XIV required reports 
on the conditions of the country from each of the general intendants. 



REPRESENTATION OF STATISTICS 271 

Prussia began in modern times the policy of making periodic 
collections of statistical data. In 1719 Frederick William I began 
requiring semiannual reports as to population, occupations, real- 
estate holdings, taxes, city finance, etc. Later these data were col- 
lected every three years. Frederick the Great also was a vigorous 
exponent of the value of statistics. He enlarged the scope of statis- 
tics in general by including nationality, age, deaths and their causes, 
conditions of agriculture, trade, manufactures, shipping, in fact, 
anything that might possibly contribute to national efficiency. 

A provision in our constitution of 1790 initiated the decennial 
census. One country after another has adopted some form of regular 
enumeration, until, in 1911, China took her first official census. 

In recent times the censuses have grown extremely elaborate. 
In 1900 the United States established a permanent Census Bureau 
whose function it is to study special problems in the light of the 
data collected and to publish the results of this study. Most leading 
nations also have special bureaus which attempt to keep the sta- 
tistics of a nation u\> to date by means of scientific estimates. An 
example of such a bureau is our National Bureau of Statistics. 
Many states have established bureaus to meet the needs of the 
state. Recently a movement has gained momentum to establish 
municipal bureaus to collect and study the data of the community 
and to instruct the public as to the significant results obtained by 
means of elaborate reports. An example of this idea is illustrated 
by the Survey Committee of the Cleveland Foundation. 



SUMMARY 

302. Chapter X has taught the meaning of the following 
words and phrases: pictogram, cartogram, bar diagrams, 
graphic curve, frequency table, class interval, central tend- 
ency, arithmetic average, mode, median, normal distri- 
bution, random sampling, compensating errors, constant 
or accumulating errors, symmetry, symmetry of a curve, 
skewness of a curve, price graph, linear equation, locus, 
axes, horizontal axis, vertical axis, ordinate, abscissa, coor- 
dinates, coordinate axes, plotting a point. 



L'7_' GKNKKAL MATHEMATICS 

303. The graphic curve may be used to show the rela- 
tion between two quantities. Specific directions were given 
showing how a graphic curve is drawn. 

304. Continuous and discrete series were illustrated and 
explained. 

305. Statistical studies are necessary to solve our social, 
governmental, and economic problems. The intelligent 
reader will profit by a knowledge of the elements of 
statistical methods. 

306. Tabulating the facts bearing on a problem in the 
form of a frequency table enables one to get a grasp on 
the problem. 

307. The word " average " as generally used may mean 
arithmetic average, mode, or median. All are measures of 
the central tendency of a mass of statistical data. The 
arithmetic average is found by figuring, the mode is found 
by inspection, and the median by counting. 

308. The law of statistical regularity was illustrated. 

309. The law of inertia of large, numbers was stated. 

310. The graph of goods purchased at a constant cost 
may be used as a " ready reckoner." 

311. The chapter has taught how to plot points on 
squared paper. 

312. The graph of a linear equation is a straight line, 
The coordinates of every point on the line satisfy the equa- 
tion, and the coordinates of every point not on the line do 
not satisfy the equation. This illustrates the locus idea. 

313. The chapter has taught the method of graphing a 
linear equation. 



CHAPTER XI 

GAINING CONTROL OF THE FORMULA; GRAPHICAL 
INTERPRETATION OF FORMULAS 

314. The formula. The formula has been defined as an 
equation which is an abbreviated translation of some prac- 
tical rule of procedure. Thus, I = Prt is a formula because 
it is an equation which is an abbreviated form of the fol- 
lowing practical rule for finding interest : To find the interest 
(expressed in dollars) multiply the principal (expressed in 
dollars) by the product of the rate (expressed hi hundredths) 
and the time (expressed in years). 

The formula is applied extensively in shop work, engi- 
neering, science, and, in fact, in every field of business and 
industry where the literature is at all technical. The sym- 
bolic form of the rule of procedure is not only more easily 
understood than the sentence form but is more easily applied 
to a particular problem. 

315. Applying the interest formula. A formula is applied 
to a problem when the known facts of the problem are sub- 
stituted in place of the letters of the formula. A formula 
may be used when all the letters except one appear as 
known facts in the problem. The pupil should study the 
following illustration : 

What is the interest on $200 at 5% for two years ? 

Solution. Substituting the known facts in the formula, 

7 = 200.^-2. 
Simplifying the right member, 

/ = $20. 
273 



274 GENERAL MATHEMATICS 

EXERCISES 

1. Find the interest on $425 at 4% for 2 yr. 

2. Find the interest on $640 at 4% for Syr. 



4.5 9 

HINT. Substitute ^ , or , for r. Why ? 

3. Find the interest on $820 at 4% for 2yr. 3 mo. 5 da. 

HINT. Reduce 2 yr. 3 mo. 5 da. to days, divide this result by 360, 
and substitute for /. Why? 

316. Other types of interest problems conveniently solved 
by special forms of I=Prt. The method of solving other 
types of interest problems is illustrated by the following 
problem : 

How much money must be invested at 5% for 2 yr. so as to 
yield $180 interest ? 

NOTE. This problem differs from Ex. 3, Art. 315, in that rate, 
time, and interest are given and the problem is to find P (the prin- 
cipal). It may be solved by substituting the three numbers given 
for the corresponding three letters of the formula. Why ? However, 
it will be found on trial to be far more convenient if we first solve 
for P in / = Prt. 

Solution. Dividing both members of the equation by rt, P. 

rt 

This may be translated into the following rule of arithmetic : 
To find the principal divide the interest by the product of the prin- 

cipal and the rate. P = is only a special form of I = Prt, but 

constitutes complete directions for finding the principal when the 
other three factors are given. 

In the proposed problem we obtain, by substituting, 



Thus the principal is $1800. 



CONTROL OF THE FORMULA 275 

EXERCISES 

1. What principal must be invested at 4|-% for 2 yr to 
yield $81? 

2. What is the principal if the rate is 6%, the time 4yr. 
3 da., and the interest $120 ? 

3. What is the rate if the principal is $500, the time 3 yr., 
and the interest $90 ? 



Here P, t, and / are given ; r is the unknown. Hence we solve 
/ = Prt for r. 

Dividing both members by P and then by t or by (Pf), 

J_ 

Pt~ 

I 

Substituting the known facts in r = , 

Pt <<\ 

1)0 6 

= - = 6%. 



500 3 100 

4. Translate r = into a rule of procedure for finding 
the rate. 

5. What is the rate if the interest is $85.50, the time 
l|-yr., and the principal $950? 

6. What is a fourth type of interest problem ? Find a 
formula most convenient for the solution of such type 
problems. 

7. Show how to obtain this formula from the form 7 = Prt. 

8. Translate into a rule of arithmetic. , 

9. Into what two parts can 1500 be divided so that the 
income of one at 6% shall equal the income of the other 
at 4% ? 

10. How can a man divide $1800 so that the income of part 
at 4% shall be the same as that of the rest at 5,% ? 

11. A certain sum invested at 4|% gave the same interest in 
2 yr. as $4000 gave in 1^ yr. at 4^ . How large was the sum ? 



276 



GENERAL MATHEMATICS 



317. Solving a formula. The process of deriving t = 

-LV 

from / = Prt is called solving the formula for t. Similarly, 
deriving the form P is called solving the formula 

/ L 

for P. The special form obtained is not only the most 
convenient form for the particular problem, but it may be 
used to solve the whole class of problems to which it 
belongs. The solving of the formulas of this chapter are 
of the practical kind and will involve little more than 
the applications of the axioms of Chapter I. 

318. Graphical illustration of interest problems. The 
relation between any two of the factors that appear in 
an interest formula may be represented graphically. 



EXERCISE 

How does the yearly interest vary on principals invested 
at 5% ? 

Substituting T 5 for r, and 1 for t, 
then /= T(j P. 

Note that this is a linear equa- 
tion involving 7 and P which may 
be plotted by the method of Art. 301. 
The table below was used to make the 
graph in Fig. 204. 

/ P 



$2.50 


$50.00 


$5.00 


$100.00 


$10.00 


$200.00 




10 15 20 25 
Interest 



Let one small unit on the horizontal 

lines represent $1 of interest, and one 

. . 

large unit on the vertical lines repre- 1N CALCULATING INTEREST ON 
sent $50 of principal invested. . PRINCIPALS INVESTED AT 5% 



. n/< n 
G. 204. GRAPH TO BE USED 



' CONTROL OF THE FORMULA 277 

Use OX as the line for plotting interests and OY for plotting prin- 
cipals. Then the point corresponding to ($2.50, $50) means 2^ small 
spaces to the right and 1 large space up. Since we know that the 
graph will be a straight line, the line OR may be safely drawn as 
soon as tw'o points are plotted. 

EXERCISES 

\. Look at the graph in Fig. 204 and tell offhand what 
interest you would expect to collect at 5% for 1 yr. on $300; 
on $350 ; on $400 ; on $60 ; on 



2. Determine by looking at the graph in Fig. 204 how much 
money you would need to invest at 5% to collect $18 interest 
in 1 yr.; $20 interest; $27.50 interest; $14 interest. 

3. How would you go about finding the interest on $12.50 
by means of a graph ? on $2000 ? 

4. Check some of the answers given by calculating the 
interest by the usual method. 

5. Graph the equation / = T ^p and use the graph to cal- 
culate interest on sums lent at 6%. 

6. Let P = $100 and r = T f 7 m the formula 1 = Prt, thus 
obtaining I = 6t. Graph 7 = 6 1 and use the graph to deter- 
mine the interest on $100 at 6% for 2 yr. ; for 2| yr. ; for 3 yr. ; 
for 4 yr. ; for 5 yr. ; for 2 mo. 

7. If possible, report in detail the methods used by your 
family banker to calculate interest. On what principles do the 
various " short cuts " rest ? 

*319. Formulas involving the amount. In the exercises 
that follow we shall study some formulas a little more diffi- 
cult to solve, but they can be understood if the funda- 
mental laws in solving equations are carefully applied. 



278 GENERAL MATHEMATICS 

EXERCISES 

1. If $400 is invested at 4%, what is the amount at the 
end of 1 yr. ? of 2 yr. ? of 3 yr. ? 

2. If $1200 is invested at 3%, what is the amount at the 
end of 2 yr. ? 

3. What is the rule for finding the amount when principal, 
rate, and time are given ? 

4. Using A for the amount and Prt for the interest, translate 
the preceding rule into a formula. 

5. The formula for the amount may also be written in the 
form A = P (1 + rt~). Prove. 

6. Solve A = P (1 + rt) for P. 

HINT. Dividing both members of the equation by the coefficient 
of P, namely, (1 + rf), we obtain P = - 

7. Translate into a rule of arithmetic the formula obtained 
in Ex. 6. 

8. Find the principal if the rate is 6%, the time 3 yr., and 
the amount $472. 



9. Find the principal if the rate is 5%, the time 3yr., and 
the amount $1150. 

10. Solve the equation .4 = 7* + Prt for t. Translate the 
resulting formula into words. 

11. Find the time if the principal is $2500, the amount 
$2725, and the rate 3%. 

12. Solve the equation A = P + Prt for r and translate the 
resulting formula into words. 

13. Find the rate if the principal is $1500, the amount 
$1740, and the time 4 yr. 

14. Summarize the advantages of solving interest problems 
by formulas. 



CONTROL OF THE FORMULA 279 

320. Evaluating a formula. The process of finding the 
arithmetical value of the literal number called for in a 
formula is called evaluating the formula. The foregoing 
exercises show that the process consists of 

1. Substituting the known numbers in the formula. 

2. Reducing the arithmetical number obtained to the simplest 
form. 

NOTE. A drill list involving these processes is given in Art. 329. 

321 . Summary of the discussion of a formula. Cultivating 
and gaining control of a formula means 

1. Analyzing an arithmetical situation so as 'to see the 
rule of procedure. 

2. Translating the rule into a formula. 

3. Solving the formula for any letter in terms of all the 
others. 

4. Evaluating the formula. 

These steps will now be illustrated in the solution of 
motion problems. We shall then proceed to solve short 
lists of exercises which should develop power in these steps. 

322. The formula applied to motion problems. In solving 
the following problems try to observe the steps summarized 
in Art. 321. 

ORAL EXERCISES 

1. If a 220-yard-dash man runs the last 50 yd. in 5 sec., at 
what rate is he finishing ? 

2. If an automobile makes 75 mi. in 2|-hr., how fast is it 
being driven ? 

3. Express the distance covered by a train in 8hr. at an 
average rate of 20 mi. per hour; of 12^ mi. per hour; of x miles 
per hour ; of x -f 3 mi. per hour. 



280 GENERAL MATHEMATICS 

4. Express the distance covered by a train in t hours at the 
rate of / miles per hour. 

5. Express the time it takes an automobile to go 150 mi. at the 
rate of 10 mi. per hour ; of 15 mi. per hour ; of 20 mi. per hour; 
of m miles per hour ; of 2 m miles per day ; of 2 m + 4 mi. per day. 

6. How long does it take to make a trip of d miles at the 
rate of r miles per hour ? 

7. The rate of a train is 30 mi. an hour. If it leaves the 
station at 1 A.M., how far away is it at 2A.M.; at 3 A.M.; at 
1 A.M.: at 5A.M.; etc.? How far away is it at 3.15A.M.; 
at 4.30A.M.; at 6.45A.M.? 

8. Denoting the distance traveled by d, find d when the rate 
is 45 mi. an hour and the number of hours is six. 

323. Distance, rate, time. The preceding exercises show 
that a problem involving motion is concerned with distance, 
rate, and time. The number of linear units passed over by 
a moving body is called the distance, and the number 
of units of distance traversed may be represented by d. 
The rate of uniform motion, that is, the number of units 
traversed in each unit of time, is called the rate (or speed) 
and is represented by r. The time, t, is expressed in 
minutes, hours, days, etc. 

ORAL EXERCISES 

1. Illustrate by familiar experiences that distance equals the 
rate multiplied by the time ; that is, that d = rt. 

2. Show how to obtain t = from d = rt. 

r 

3. Translate t = into a rule for finding the time. 

4. Show how to obtain r = - from d = rt. 

5. Translate r = - into a rule for finding the rate. 



CONTROL OF THE FORMULA 281 

6. A motor cycle goes 110 mi. in 5 hr. and 30 min. Assum- 
ing the rate to be uniform, what is the rate ? Which of the 
formulas did you use ? 

7. Sound travels about 1080 ft. per second. If the sound of 
a stroke of lightning is heard 2.5 sec. after the flash, how far 
away is the stroke ? Which form of the formula is used ? 

8. How many seconds would it take the sound to reach the 
ear if a tree 2376 ft. distant were struck by lightning ? Which 
form of the formula is used ? 

WRITTEN EXERCISES 

1. A motor boat starts 10 mi. behind a sailboat and runs 
14 mi. per hour, while the sailboat makes 6 mi. per hour. How 
long will it require the motor boat to overtake the sailboat ? 

Let x be the number of hours it takes the motor boat to overtake 
the sailboat. 

Then, according to the data of the problem : 

for the motor boat, for the sailboat, 

t = x. t = x. 

r = 14. r = 6. 

Hence d = 14 x. Hence d 6 x. 

Since the motor boat must go 10 mi. more than the sailboat, the 
following equation expresses the conditions of the problem : 
14 x = 6 x + 10. 

Solve the equation to find the value of x, which turns out to 
be llhr. 

2. A and B live 22^ mi. from each other. In order to meet 
A, B leaves home an hour earlier than A. If A travels at the 
rate of 4 mi. an hour and B at the rate of 3| mi. an hour, when 
and where will they meet ? 

3. A northbound and a southbound train leave Chicago at 
the same time, the former running 4 mi. an hour faster than 
the latter. If at the end of 1^ hr. the trains are 126 mi. apart, 
find the rate of each. 



282 GENERAL MATHEMATICS 

4. In running 280 mi. a freight train whose rate is | that 
of an express train takes 3-|- hr. longer than the express train. 
Find the rate of each. 

5. An automobile runs 10 mi. an hour faster than a 
motor cycle, and it takes the automobile 2 hr. longer to run 
150 mi. than it takes the motor cycle to run 60 mi. Find the 
rate of each. 

6. A man rows downstream at the rate of 8 mi. an hour 
and returns at the rate of 5 mi. an hour. How far downstream 
can he go and return if he has 5| hr. at his disposal ? At what 
rate does the stream flow ? 

7. Chicago and Cincinnati are about 250 mi. apart. Suppose 
that a train starts from each city toward the other, one at the 
rate of 30 mi. per hour and the other at the rate of 35 mi. per 
hour. How soon will they meet ? 

8. A train is traveling at the rate of 30 mi. an hour. In 
how many hours will a second train overtake the first if the 
second starts 3 hr. later than the first and travels at the rate 
of 35 mi. an hour ? 

*9. A and B run a mile race. A runs 20 ft. per second, and 
B 19^ ft. per second. B has a start of 32 yd. In how many 
seconds will A overtake B ? Which will win the race ? 

* 10. A bullet going 1500 ft. per second is heard to strike 
the target 3 sec. after it is fired. How far away is the target ? 
(Sound travels at the rate of about 1080 ft. per second.) 

11. A motor cyclist rode 85 mi. in 5 hr. Part of the distance 
was on a country road at a speed of 20 mi. an hour and the 
rest within the city limits at 10 mi. an hour. Find how many 
hours of his ride were in the country. 

12. Two boats 149 mi. apart approach each other, leaving at 
the same time. One goes 10 mi. per hour faster than the other, 
and they meet in 2 hr. What is the rate of each ? 



CONTROL OF THE FORMULA 



283 



30 



45 



60 



324. Graphical illustration of a motion problem. Many 
motion problems can be conveniently illustrated graphi- 
cally, as the student will discover if he solves the following 
exercises. 

EXERCISES 

1. In the Indianapolis races De Palma drove his car at a 
rate varying but little from 90 mi. per hour. Draw a graph 
showing the relation between the distance and d t 
time of De Palma's performance. 

Substituting 90 in d = tr, 
d = 9Qt. 

Note that d 90 1 is a linear equation which 
may be graphed (see table and Fig. 205). Ten small 
units on the vertical axis represent 30 mi.; ten small units on the 

horizontal axis represent ^ hr. 



2. Determine from the graph in Fig. 205 how many miles 

De Palma drove in 2 hr. ; in 1^ hr. ; in 1 hr. 24 min. ; in 40 min. ; 
in 4 min.; in 2 hr. 12 min. 

3. Determine by the graph 
in Fig. 205 how long it took 
De Palma to go 50 mi. ; 40 mi. ; 
GO mi.; 75 mi.; 140 mi.; IGOmi.; 
10 mi. 

Obviously the preceding re- % 
suits could be calculated either 
by arithmetic or by the formula. 
However, the graph has the advan- 
tage of revealing all the results 
in vivid fashion. 




i l 1J 
Time in Hours 



FIG. 205. THE GRAPH OF A MOTION- 
PROBLEM FORMULA 



4. Draw a graph showing 
the distances traversed by a passenger train running uniformly 
at the rate of 40 mi. per hour for the first ten hours of its trip. 
*5. Find out, if possible, what use railroad officials make of 
graphs in arranging schedules. 



284 GENERAL MATHEMATICS 

325. Circular motion. Circular motion is of frequent 
occurrence in mechanics. A familiar illustration is found 
in the movement of the hands of a clock. 

EXERCISES 

1. At what time between 3 and 4 o'clock are the hands of 
a clock together ? h 

Solution. Let x (Fig. 206) equal the 
number of minutes after 3 o'clock when 
the hands are together ; that is, x equals 
the number of minute spaces over which 
the minute hand passes from 3 o'clock 
until it overtakes the hour hand. 

x 
Then equals the number of minute 

FIG. 206. CLOCK PROBLEMS 
spaces passed over by the hour hand. ILLUSTRATE A TYPE OF ClR . 

Why? CULAK MOTION 

Since the number of minute spaces 

from 12 to 3 is 15, and since the whole is equal to the sum of its 
parts, it follows that 




Whence x = 16 ^ 4 T min. 

Therefore the hands are together at 16^ min. after 3 o'clock. 

2. At what time between 4 and 5 o'clock are the hands of 
a clock together ? 

HINT. Draw a figure similar to the one for Ex. 1. 

Notice that the formula for a clock problem is x = + m, 

\.t 

where m equals the number of minute spaces the minute hand must 
gain in order to reach the desired position. 

3. At what time between 2 and 3 o'clock are the hands of 
a clock 15 min. apart ? 

HINT. Draw a figure, think the problem through, and then try 
to see how the formula in Ex. 2 applies. 



CONTROL OF THE FORMULA 285 

4. At what time between 2 and 3 o'clock are the hands of 
a clock 30 minute spaces apart ? 

5. What angle is formed by the hands of a clock at 2.30? 
*6. At what time between 5 ,and 6 o'clock are the hands of 

a clock 20 min. apart ? How many answers ? How may these 
results be obtained from the formula of Ex. 2 ? 

326. Work problems. The work problem is another 
type of problem easily solved by formula. 

EXERCISES 

1. One pipe will fill a tank in 3 hr. and a second pipe can 
fill it in 4 hr. How long will it take to fill the tank if both 
pipes are left running ? 

Let n = the number of hours it will take both pipes to fill the tank. 

Then = the part of the tank filled in 1 hr., 
n 

i = the part of the tank filled by the first pipe in 1 hr., 
and ^ = the part of the tank filled by the second pipe. 

Hence - +'- = - Why? 

3 4 n 

Multiplying by 12 n, 4 n + 3 n = 12, 
or 7 n = 12. 

Whence n = If hr. 

2. A can lay a drain in 5 da. and B can do it in 7 da. How 
long will it take both working together ? 

3. One boy can drive his car over a trip in 8 hr. and a 
second boy can make the trip in 5 hr. How long would it take 
them to meet if each started at an end ? 

NOTE. It is clear from the foregoing problems that any numbers 
would be used just as 3 and 4 are used in Ex. 1. Hence a formula 
may be obtained, as is shown by Ex. 4. 



286 GENERAL MATHEMATICS 

4. A can do a piece of work in a days and B can do it in 
b days. How long will it take them to do it together ? 

Let n = the number of days it will take 

them together. 

Then - = the amount of work they can do 

n , , 

in 1 da., 

- = the amount A can do in 1 da., 
a 

and - = the amount B can do in 1 da. 

b 

1,1 1 
Hence _ + _ _. 

a b n 

Multiplying by aim, l>n + an = ah, 
(J> + a)n ah, 
ab 



a + I > 
NOTE. Any problem of the type of Ex. 1 on page 285 may be 

solved by using the equation n = as a formula. Thus, to solve 

a + b 

Ex. 1 let a = 3, b 4. Then n = = = 1^ hr. 

5. One boy can make a paper route in 2 hr. and his friend 
can make the route in 1^ hr. How long will it take the two 
together ? (Solve by formula.) 

*6. Suppose that in Ex. 1 on page 285 the second pipe is an 
emptying pipe, how long will it take to fill the tank if both 
pipes are running ? What form does the formula take ? 

*7. A can sweep a walk in 7 min., B in 8 min., and C in 
10 min. How long will it take them working together ? What 
form does the formula for a work problem take ? 

*8. A could lay a sidewalk in 3 da., B in 4 da., and C in 
4.5 da. How long does it take them when working together ? 
Solve by substituting in the formula for Ex. 7. 



CONTROL OF THE FORMULA 287 

327. Translating rules of procedure into formulas. Write 
each of the following in the form of a formula : 

1. The area of a triangle equals the product of half the 
base times the altitude. 

2. The area of a rectangle equals the product of its base 
and altitude. 

3. The area of a parallelogram equals the product of its 
base and altitude. 

4. The area of a trapezoid equals one half the sum of the 
parallel bases multiplied by the altitude. 

5. The volume of a pyramid equals one third the base 
times the altitude. 

6. The length of a circle is approximately equal to twenty- 
two sevenths of the diameter. 

7. The circumference of a circle is equal to TT times the 
diameter. 

8. The area of a circle is TT times the square of the radius. 

9. The product equals the multiplicand times the multiplier. 

10. The product obtained by multiplying a fraction by a 
whole number is the product of the whole number and the 
numerator divided by the denominator. 

11. The quotient of two fractions equals the dividend multi- 
plied by the inverted divisor. 

12. The square root of a fraction equals the square root of 
the numerator divided by the square root of the denominator. 

13. The square of a fraction is the square of the numerator 
divided by the square of the denominator. 

14. The rule for calculating the cost of one article when 
you know that a certain number of them cost so much ; write 
the cost of m articles. 

15. The rule for expressing years, mouths, and days as years. 



288 GENERAL MATHEMATICS 

16. The rule for calculating the area of three adjacent rooms 
of different lengths but the same width. 

17. The rule for calculating the area of the floor of a 
square room. 

18. The rule for finding the cost of a telegram. 

19 . The rule for finding the area of a figure cut from cardboard, 
given its weight and the weight of a square unit of cardboard. 

20. The rule for finding the amount of available air for each 
person in a classroom, given the dimensions of the room and 
the number in the class. 

21. The rule for finding the weight of a single lead shot, 
given the weight of a beaker with a given number of shot in 
it and the weight of the empty beaker. 

22. The rule for predicting the population of a town after 
a certain number of months, given the present population, the 
average number of births, and the average number of deaths. 

23. The rule for finding the distance apart, after a given 
time, of two cars which start from the same point and travel 
in opposite directions at different speeds. 

24. The same as in Ex. 23 except that the cars are m miles 
apart at starting. 

25. The same as in Ex. 23 except that the cars go in the 
same direction with different speeds. 

26. The reading on a Fahrenheit thermometer is always 32 
greater than ^ of the reading on a centigrade thermometer. 

27. The reading of a centigrade thermometer may be cal- 
culated by noting the reading on the Fahrenheit, subtracting 
32, and taking | of this result. 

328. Graphic representation of the relation between the 
readings on centigrade and Fahrenheit thermometers. The 
last two exercises deal with two types of thermometers 
that are used to measure temperature. Fig. 207 shows that 



CONTROL OF THE FORMULA 



289 



F 



Boiling 100' 
90 f 
80' 
70' 
60' 
50' 
40' 
30' 
20' 
10' 

Freezing 0' 

-10' 

-17.78' 



the only fundamental difference is the different graduations 
of the scale. In the Fahrenheit thermometer the place 
where the mercury stands if immersed in freezing water is 
32 ; on the centigrade it is zero. This defines the freezing 
point on each. The boiling point is 
marked 212 on the Fahrenheit and 
only 100 011 the centigrade. Hence 
the Fahrenheit thermometer has 180 
divisions of the scale in the interval 
from freezing to boiling, while the 
centigrade has but 100. This means 
that a unit on the centigrade is 
longer, or for any space on the centi- 
grade there are i|$ times, or | times, 
as many Fahrenheit units. Hence 
the number of units in a Fahrenheit 
reading equals -| times as many cen- 
tigrade units plus 32 (which are 
below the freezing point). Stated 
as a formula, F=fC + 32. 

We could, of course, always trans- BETWEEN THE CENTI- 

late the reading on one thermometer GRAI)E AND FAHRENHEIT 

... ,. THERMOMETERS 

to the corresponding reading on the 

other by means of this formula. It is far easier, how- 
ever, to graph the linear equation F = |C + 32, which 
will reveal possible relations, as is shown in Fig. 208. 

NOTE. The pupil should construct the graph independently of 
the text, using a table similar to the following : 

C. F. 

Let C.= 0, then F. = 32. 

Let C. = 25, then F. = 77. 

Let C. = 15, then F. = 59. 



212 

194 

-176 

-158 

140 

-122 

-104 

86 

68 

50 

32 

14 





FIG. 207. THE RELATION 




290 



CKNKRAL MATHEMATICS 



EXERCISES 

1. Determine by the graph the corresponding Fahrenheit 
readings for the following centigrade readings : 5, 10, 20, 
30, - 5, - 10, - 15, - 25. 

2. Determine by the graph the centigrade readings cor- 
responding to the following Fahrenheit readings : 80, 70, 
60, 30, 20, 10, - 5, - 10. 

3. In the formula F = f C 
-f- 32 substitute in each case 
the two numbers you think are 
corresponding readings. The 
error should be very small. 

4. Normal room tempera- 
ture is 68 F. What is it 
centigrade ? 

5. The normal temperature 
of the human body is 98.4 F. 
What is it centigrade ? 

6. What temperature centi- 
grade corresponds to F.? 



7. Could you go skating at FlG 20g A GRAPH T0 BE rSK1) JN 

15 C. ? CHANGING CENTIGRADE READINGS 

TO FAHRENHEIT AND VICE VERSA 

8. In your general-science 

course you are told that mercury freezes at 40 F. What 
is this centigrade ? 

9. Would your classroom be comfortable at 25 C. ? 

329. Evaluating a formula. Find the value of the letter 
called for in each of the exercises given on page 291. 
When no explanation is given, it is assumed that the 
student recognizes the formula. 




CONTROL OF THE FORMULA 



291 



EXERCISES 

1. Given C = | (F - 32). Find C. if F. = 0; 32; 
212; 100. 

2. Given F = |C.+ 32. Find F. if C. = 0; 100; 
-20; 60. 

3. Given d = rt. Find d if / = 87.5 mi. per hour and 
t = 12 hr. ; if r = 10^ ft. per second and t 10 sec. 

4. Given r = - Find r if d = 1 mi. and t = 4 min. 16 sec. ; 

{/ 

if d = ^ mi. and t = 2.07 sec. 

5. Given A=P+ Prt. Find A if P = $240, r = 4-}%, and 
*=lyr. 2 mo. 3 da.; if P = $128, r=6%, and = 2yr. 3 da.; 
if P = $511, r = 6%, and t = 20 yr. 

6. Given V=lwh. Find F (see Fig. 209) 
if I =12.2 ft., w = 8.3 ft., and h = 6.4 ft.; 
if = 9.3 in., iv = 5.6 in., and h = 1 in. 



7. 



Find Tiff =63 ft., 




FIG. 209. RECTAX- 

w = 2.4 ft., and h = 1.6 ft.; if I 3 cm., GULAR PARALLELE- 
iv = 2.1 cm., and h = 1.4 cm. PIPED 

8. Given c = ^- d. Find c if d = 1 f t. ; 1 in. ; 4 in. ; 10 in. ; 
5-1 in. 

9. Given A = Trr 2 (TT = - 2 T 2 -). Find .4 if r = lin.; 5ft.; 
10 yd. ; 7 m. ; 8.5 cm. 

10. The volume F of any prism (Fig. 210) 
is equal to the product of its base B and 
its altitude h. Find V if B = 246.12 sq. in. 
and h = 12 in.; if B = 212.44 sq. in. and 
& = 2| ft. 

11. The lateral surface L of a right 

prism equals the perimeter of the base FlG 2 io. PRISM 

P times the altitude h. Find L if P = 

126 in. and h = 11 in.; if P = 21.6 in. and h = 0.35 in. 




292 



GENERAL MATHEMATICS 



CYLINDER 



12. The volume of a right cylinder (Fig. 211) is equal to 
the product of its base and height. The formula is V = TrrVi, 
where r is the radius of the circular base. Find Vii r=10.2 cm. 
and h = 9 cm. ; if r = 6 in. and h 12 in. 

13. The lateral surface of a right cylinder 
equals the product of the altitude and the 
circumference of the base. The formula 
usually given is S = Ch. Find S if C = *f ^ in. 
and h = 10 in. 

14. The entire surface T of a right cylinder 
equals the circumference of the circular base 

times the sum of the altitude and the radius of Fj( . 2 n 
the base ; that is, T = 2 irr(r + h). Find T if 
A =10 in. and r = 5 in.; if A = 2ft. and r=l ft. 

15. The volume V of any pyramid (Fig. 212) 
is equal to one third the product of its base 

B and its altitude h ; that is, V = Find 

o 

V if = 200sq. in> and A =12 in.; if B = 
24.6 sq. in. and h = 2 ft. 

16. The lateral area S of a regular pyramid 

is equal to one half the product of the perimeter P of its base 

Pi 

and its slant height ; that is, S = Find A if P = 10. 6 in. 

t 

and I = 8.2 in. ; if P = 4.3 cm. and I = 15 cm. 

17. The lateral area S of a right circular 
cone (Fig. 213) is equal to one half the prod- 
uct of its slant height I and the circumfer- 
ence C of its base. Write the formula for S, 
and find S if 1= 14.6 in. and C = 10in. ; if 

I = 3.6 ft. and C = 31.416 ft. FIG. 213. RIGHT 

18. The lateral area S of a right circular cone ClRCULAR CONE 
is Trrh, where r is the radius of the base and h is the slant height. 
Find S if r = 10 in. and h = 10 in. ; if r = 10 in. and h = 26.2 in. 




FIG. 212. PYRAMID 




CONTROL OF THE FORMULA 293 

19. The entire surface T of a right circular cone equals the 
lateral area I plus the area of the base ; that is, T = trrl + Trr 2 , 
or 7rr(l + r). Find T if I = 10 in. and r = 5 in. ; if I = 12.6 in. 
and r = 6 in. 

20. An object falling from rest falls in a given time a 
distance equal to the product of 16 and the square of the 
number of seconds it has fallen; that is, d = 16t 2 . Find 
d if t = 1 sec. ; 2 sec. ; 3 sec. ; 4 sec. 

21. An object thrown downward travels in a given time a dis- 
tance equal to the product of 16 and the square of the number 
of seconds it has fallen, plus the product of the velocity with 
which it is thrown and the number of seconds it is falling. The 
formula is S = 16 1* + Vt. Find S if t = 3 sec. and V= 13 ft. per 
second; if t = 5 sec. and F=100ft. per second. 

22. The volume of a sphere (Fig. 214) equals 
the cube of the radius multiplied by ^ TT. Find V 
if y = 1 in.; 10 in.; 5 in.; 10ft.; 12ft. 

23. The surface S of a sphere is equal to FIG. 214. THF, 
4 7H- 2 . Find S if r = 10 in. ; 12 ft. ; 6j ft. 

*24. The force of pressure P of the wind, in pounds per 
square foot, is given by the equation P = 0.005 F 2 , where V is 
the velocity pf the wind in miles per hour. What would be the 
total pressure of this wind against the side of a wall 25 ft. 
high and 80 ft. long of a wind blowing 30 mi. per hour ? 

*25. Show that the formula for the length I of a belt passing 
around two pulleys of the same size whose radii are each r feet, 
and the distance between whose centers is d feet, is 1= 2 7rr+ 2 d. 
Find I when r = l and d = 4|. 

26. In a price list the cost of sewer pipe per foot of length 
is given by the formula C = 0.4 d 2 + 14, where d is the diam- 
eter of the pipe in inches and C the cost in cents. What will 
be the cost of 20 ft. of pipe 2 in. in diameter ? 




294 GENERAL MATHEMATICS 

330. Practice in solving for any letter. It is often desir- 
able to solve a formula for some particular letter in that 
formula. Too often the student will recognize a formula 
provided it stands in the form in which it is commonly 
written, but will not appreciate its meaning if it is written 
in a different way. For example, how many students 

V 

would recognize the formula c = as the well-known 

formula V=abc? It is the same formula except that it 
is in a different form. If the student realizes this, it helps 
him to gain control of the formula. The following exer- 
cises will furnish practice in solving for particular letters. 

EXERCISES 

Solve each of the following formulas for the letter or letters 
indicated : 

ab Bh 

1. .-I = for a ; for />. 11. T = for h. 

2. r = aJcforc; fora; forb. Pi. 

12. .1 = for /. 

3. <? = rtforr; for t. 

D 13. A = P + Prt for t. 

4. ^ = Y for,/;for/>. 14 . r = |(F- 32) for F. 

5. WA*WJJ*to**W t 15 Sss9 * fmi ,. lory. 



6- V- for w; for A. i 6 . s = 2 irrh + 2 ir>* for h. 

7. C = 2 . 5 (7 for r/. 17. .1 = l 9 ' 2 for // ; 

8. r = 41 ,. for r. for ^ ; for 6 a . 

9. V = Bh for B. 18 . c = _A_ for E . for ; , . 
10. V=*i*h for A. for r. 



CONTROL OF THE FORMULA 295 

SUMMARY 

331. This chapter has taught the meaning of the following 
words and phrases : formula, solving a formula, evaluating 
a formula, applying a formula, centigrade, Fahrenheit. 

332. A formula is a conveniently abbreviated -form of 
some practical rule of procedure. 

333. A clear understanding of a formula implies: 

1. An analysis of some arithmetical situation so as to 
arrive at some rule of procedure. 

2. Translating the rule into a formula. 

3. The ability to solve for any letter in terms of the 
other letters in the formula. 

4. The ability to apply the formula to a particular 
problem and to evaluate the formula. 

334. The preceding steps were illustrated in detail by 
applications to interest problems, to problems involving 
motion, to work problems, to thermometer problems, and 
to geometric problems. 

335. The graphical interpretations suggested economical 
methods of manipulating a formula. For example : 

1. Simple-interest problems were solved by the formulas 
1 = Prt and A = P + Prt. 

2. A problem involving uniform motion in a straight 
line was solved by the formula d = rt. 

3. The relation between centigrade and Fahrenheit 
readings was expressed by the formula (7 =!(/* 32). 

336. While the important thing in this chapter is the 
power of manipulating and evaluating a formula, the stu- 
dent was given the meaning of most of the formulas in 
order to have him realize from the very outset that both the 
formulas and their manipulation refer to actual situations. 



296 GENERAL MATHEMATICS 

HISTORICAL NOTE. The development of the formula belongs to 
a very late stage in the development of mathematics. It requires a 
much higher form of thinking to see that the area of any triangle 

can be expressed by A = than to find the area of a particular 

A 

lot whose base is two hundred feet and whose altitude is fifty feet. 
Hence, it was very late in the race's development that letters were 
used in expressing rules. 

The early mathematicians represented the unknown by some 'word 
like res (meaning " the thing "). Later, calculators used a single letter 
for the unknown, but the problems still dealt with particular cases. 
Diophantus, representing Greek mathematics, stated some problems 
in general terms, but usually solved the problems by taking special 
cases. Vieta used capital letters (consonants and vowels) to represent 
known and unknown numbers respectively. Newton is said to be 
the first to let a letter stand for negative as well as positive numbers, 
which greatly decreases the number of formulas necessary. 

While the race has had a difficult time discovering and under- 
standing formulas, it takes comparatively little intelligence to use a 
formula. Many men in the industrial world do their work efficiently 
by the means of a formula whose derivation and meaning they do 
not understand. It is said that even among college-trained engineers 
only a few out of every hundred do more than follow formulas or 
other directions blindly. Thus, it appears that for the great majority 
only the immediately practical is valuable. However, we can be 
reasonably sure that no one can rise to be a leader in any field by 
his own ability without understanding the theoretical as well as 
the practical. 

The formula is very important in the present complex industrial 
age. A considerable portion of the necessary calculation is done by 
following the directions of some formula. Therefore to meet this 
need the study of the formula should be emphasized. In discussing 
the kind of mathematics that should be required Professor A. R. 
Crathorne (School and Society, July 7, 1917, p. 14) says: "Great 
emphasis would be placed on the formula, and all sorts of formulas 
could be brought in. The popular science magazines, the trade 
journals and catalogues, are mines of information about which the 
modern boy or girl understands. The pupil should think of the 
formula as an algebraic declarative sentence that can be translated 




ARCHIMEDES (287-212 B.C.) 
(Bust in Naples Museum) 



298 GENERAL MATHEMATICS 

into English. The evaluation leads up to the tabular presentation 
of the formula. Mechanical ability in the manipulation of symbols 
should be encouraged through inversion of the formula, or what the 
Englishman calls 'changing the subject of the formula.' We have 
here also the beginning of the equation when our declarative sentence 
is changed to the interrogative." 

Archimedes (287-212 H.C.), a great mathematician who studied 
in the university at Alexandria and lived iu Sicily, loved science 
so much that he held it undesirable to apply his information to 
practical use. But so great was his mechanical ability that when 
a difficulty had to be overcome the government often called on 
him. He introduced many inventions into the everyday lives of 
the people. 

His life is exceedingly interesting 1 . Read the stones of his detec- 
tion of the dishonest goldsmith ; of the use of burning-glasses to 
destroy the ships of the attacking Roman squadron ; of his clever 
use of a lever device for helping out Hiero, who had built a ship 
so large that he could not launch it off the slips ; of his screw for 
pumping water out of ships and for irrigating the Nile valley. He 
devised the catapults which held the Roman attack for three years. 
These were so constructed that the range was either long or short 
and so that they could be discharged through a small loophole 
without exposing the men to the fire of the enemy. 

When the Romans finally captured the city Archimedes was killed, 
though contrary to the orders of Marcellus, the general in charge of 
the siege. It is said that soldiers entered Archimedes' study while he 
was studying a geometrical figure which he had drawn in sand on 
the floor. Archimedes told a soldier to get off the diagram and not 
to spoil it. The soldier, being insulted at having orders given to 
him and not knowing the old man, killed him. 

The Romans erected a splendid tomb with the figure of a sphere 
engraved on it. Archimedes had requested this to commemorate his 
discovery of the two formulas : the volume of a sphere equals two- 
thirds that of the circumscribing right cylinder, and the surface of 
a sphere equals four times the area of a great circle. You may also 
read an interesting account by Cicero of his successful efforts to 
find Archimedes' tomb. It will be profitable if the student will 
read Ball's " A Short History of Mathematics," pp. 65-77. 



CHAPTER XII 

FUNCTION ; LINEAR FUNCTIONS ; THE RELATED IDEAS OF 

FUNCTION, EQUATION, AND FORMULA INTERPRETED 

GRAPHICALLY; VARIATION 

337. Function the dependence of one quantity upon 
another. One of the most common notions in our lives is 
the notion of the dependence of one thing upon another. 
We shall here study the mathematics of such dependence 
by considering several concrete examples. 

EXERCISES 

1. Upon what does the cost of 10yd. of cloth depend? 

2. If Resta drives his car at an average rate of 98.3 mi. per 
hour, upon what does the length (distance) of the race depend ? 

3. A boy rides a motor cycle for two hours. Upon what 
does the length of his trip depend ? 

4. How much interest would you expect to collect in a 
year on $200? 

5. Upon what does the length of a circular running track 
depend ? 

6. A man wishes to buy wire fencing to inclose a square lot. 
How much fencing must he buy ? 

7. State upon what quantities each of the following depends : 

(a) The amount of sirloin steak that can be bought for a dollar. 

(b) The number of theater tickets that can be bought for 
a dollar. 

(c) The height of a maple tree that averages a growth of 
4 ft. per year. 

299 



300 GENERAL MATHEMATICS 

(d) The time it takes you to get your mathematics lesson if 
you solve one problem every three minutes. 

(e) The value of a submarine as a merchant vessel. 

(f) The rate of interest charged by your local bank. 

(g) The perimeter 4 x 4 of the rectangle in Fig. 215. 

The preceding exercises illustrate the dependence of one 
quantity upon another. We have had numerous other 
examples of dependence in the chapters on statistics and 
the formula. In fact, every practical formula implies that 
the value of some quantity depends upon one or more 
others. Thus the circumference of a circular running track 
depends upon the diameter. When a quantity depends 
upon another quantity for its value, it is said to be a func- 
tion of the latter. Thus the area of a circle is a function 
of the diameter because it depends upon the diameter for 
its value ; the amount of sirloin steak that can be bought 
for a dollar is a function of the price per pound ; and the 
expression 4 x 4 is a function of x because its value 
changes with every change in the value of x. 

See if you can illustrate the idea of function by ten 
familiar examples not given above. 

338. Variable. A number that may change, assuming a 
series of values throughout a discussion, is called a variable. 
It is not obliged to vary it is " able to vary." Thus the price 
of wheat and the number s in the equation A = s 2 are variables. 

339. Dependent and independent variables. In the formula 
C = ird the number d is said to be the independent variable. 
In a discussion or in the construction of circles we may 
take its value equal to any number we please. On the 
other hand, the value of C is automatically fixed once the 
value of d is determined. Because of this fact C is called 
the dependent variable. 



FUNCTION INTERPRETED GRAPHICALLY 301 

EXERCISES 

1. What is the value of C in the equation C ird if d 2? 
if d = 5? if d = W? 

2. Illustrate the ideas of dependent and independent vari- 
ables with examples chosen from the text or from your own 
experience. 

340. Constant. The number TT in the formula C = Trd 
differs from C and d inasmuch as it never changes at any 
time in the discussion of circles. This number is approxi- 
mately -2y 2 -, or 3.1416, whether we are dealing with small or 
large circles. We therefore call a number like this, which 
has a fixed value, a constant. Obviously any arithmetical 
number appearing in a formula is a constant ; thus the 2 in 

A = and the - and the 32 in .7^= ^ 7+ 32 are constants. 
2 

EXERCISE 

Turn to Chapter XI, on the formula, and find five formulas 
that illustrate the idea of a constant. 

341. Graph of a function. A graph may be constructed 
showing how a function changes as the value of the 
independent variable changes. x 

The rectangle in Fig. 215 is a 

picture (either enlarged or re- x-z x-z 

duced) of every rectangle whose 
length exceeds its width by two 
units. We shall now proceed to FlG 2 i5 

show graphically that the perim- 
eter varies with every change in the value of x. The table 
on the following page gives the corresponding values for 
the length x and for 4 x 4, the perimeter. 



302 



GENERAL MATHEMATICS 



X 


3 


4 


5 


6 


7 


8 


9 


10 


4x-4 


8 


12 


16 


20 


24 









4x-4 



--20 



If we plot the points corresponding to (3, 8), (4, 12), 
(5, 16), etc., using the horizontal axis to plot the values of x 
and the vertical axis to plot the values of 4 x 4, we obtain 
the points as shown on the straight line AB in Fig. 216. The 
line shows that as x increases, the value of 4 x 4 increases 

accordingly. 

EXERCISES 

1. Tell in your own words how the graph in Fig. 216 
shows that the function 4 x 4 increases as x increases. 

2. Determine from the graph 
the perimeters of . rectangles 
whose lengths are as follows : 
8 in.;, 9 in.; 10.5 in. ; 11 in. 

3. Determine from the graph 
the length of the rectangles 
whose perimeters are as follows : 
30 in.; 25 in.; 20 in. ; 18 in.; 
10 in. ; 3 in. ; in. 

4. Suppose you chose to make 
a particular rectangle 10 in. long. 
How long would the perimeter be? 
How does the graph show this ? 

5. How long would you make 
a rectangle of the same shape 
as the one in Fig. 215 so as to 
have its perimeter 16 in. ? How 
does the graph show this ? 

6. Relying on your past experience, tell how many rectangles 
you could construct in the shop or construct in your notebook 
"whose length shall exceed their width by two units ". 



FIG. 216. GRAPH SHOWING THAT 
THE PERIMETER OF THE RECTAN- 
GLE IN FIG. 215 is A FUNCTION 
OF THE LENGTH 



FUNCTION INTEEPBETED (GRAPHICALLY 303 

342. Linear function. Since the graph of the expression 
4 x 4 is a straight line, the function is called a linear func- 
tion. If we let y represent the value of the linear function, 
we get the corresponding linear equation y 4 jc 4. 

EXERCISE 

Give five examples of linear functions. 

343. Solving a family of equations by means of the 
graph. The' graph of the function 4 x 4 may be used 
to solve all equations one of whose members is 4 r 4 
and the other some arithmetical number or constant. For 
example, if in the equation y = 4x 4 we let ;z/ = 16, 
then the equation 4 # 4 = 16 may be interpreted as rais- 
ing the question, What is the value 'of x that will make 
4. r 4=16? In order to answer this question we find 
16 on the ?/-axis (the vertical axis), pass horizontally to 
the graph of 4 x 4, and read the corresponding value 
of x. The corresponding value of x is seen to be 5. Hence 
4 x 4 = 16 when x = 5. 

As a verbal problem the equation 4 a* 4=16 may be 
translated into the following interrogative sentence: What 
shall be the length of the rectangle in order that it may 
have a perimeter of 16 ? A glance at the graph is sufficient 
to determine the answer ; namely, 5. 

EXERCISES 
Solve by graph, and check the following equations : 

1. 4z-4 = 20. 5. 4z-8 = 2. 

HINT. Add 4 to both mem- 

2. 4ic 4 = Z4. . 

bers so as to obtain the equation 

3. 4.r - 4 = 12. 4z-4 = 6. 

4. 4- 4 = 6. 6. 4x-5 = 13. 



304 <;KM-:KAL MATHEMATICS 

7. 4.r-9 = 10. 9. 4ir+ 2 = 12. 

8. 4.r + 6 = 26. 10. 4a: + 5 = 19. 

HIM. Subtract 10 from both - - J. _i_ 1 <" _ <><i 
members so as to obtain the 

result 4 x - 4 = 16. 12. 4 ./ - 4 = 0. 

344. The graphical solution of the function set equal to 
zero. Problem 12, Art. 343, is an interesting special case 
for two reasons: (1) It gives us an easy method of find- 
ing the value of x in the equation 4^ 4 = 0. We need 
only refer to the graph and observe where the line crosses 
the o>axis. The line is seen to cross where # = 1. This value 
of x checks because 4-14 = 0. Hence x = 1 is a solu- 
tion of the equation 4 x 4 = 0. (2) It furnishes us a 
graphic method for solving all linear equations in one 
unknown because every equation in one unknown can be 
thrown into a form similar to 4 x 4 = 0. Show how this 
may be done with the equation 3:r + 7 = :r + 12. 

EXERCISES 

1. Solve graphically the equation 3x + 7 = x + 12. 

HINT. The equation may be written in the form 2 x 5 = 0. Why ? 

Graph the function 2x 5 just as we graphed 4# 4 (Fig. 216). 

See where the graph of 2 x 5 crosses the x-axis. This is the 
correct value of x. 

Check by substituting this value of x in the original equation 
3 x + 7 = x + 12. 

2. Solve the following linear equations by the graph, and 
check the results : 

(a) ox + 2 = 2x + S. (e) 2.ox + 9 = 3x + 7. 

(b) 6x - 5 = 4x + 2. (f) |_ 7 = x - 5. 
(c)5a; + 8 = 8x-4. x x 

(d) 11 x - 9 = 14* + 7. ') 3 4 = 2~ 7 ' 



FUNCTION INTERPRETED GRAPHICALLY 305 



HISTORICAL NOTE. Perhaps the most important idea which a 
student can learn from an elementary study of mathematics is the idea 
of a function. This is given far greater prominence in European than 
in American texts. Indeed, it is remarkable that though the function 
idea is generally admitted to be a fundamental notion in mathematics, 
the teaching of the notion is often neglected. 

From the simple illustrations in this text it will appear that many 
familiar facts and principles of natural phenomena can be expressed 
as functions. A study of these facts in the form of a mathematical 
function throws much additional light on them. Thus, a clear 
understanding of the principles of light, sound, and electricity could 
not be obtained without a study of the mathematical functions by 
which these principles are expressed. The dependence of one quantity 
upon another is one of the fundamental notions of human experience. 
It is valuable to learn to treat this notion mathematically. 

Teachers of mathematics will also remember Professor Felix Klein 
as one who has improved mathematics teaching by insisting that it 
be made more meaningful to pupils of high-school age by introducing 
and emphasizing the notion of a function. 

345. Direct variation. If a man walks at the rate of 
3 mi. an hour, the distance d which he walks in a given 
time t is found from the equation d = 3 t. The following 
table gives the distance passed over in 1 hr., 2 hr., 3 hr., etc. 
and shows 

1. That the greater the time, the greater the distance 
passed over. 

2. That the ratio of any time to the corresponding 
distance is I. 



Number of hours . . 


1 


2 


3 


4 


5 


6 


7 


8 


Distance passed over 


3 


6 


9 


12 


15 


18 


21 


24 



When two numbers vary so that one depends upon the 
other for its value, keeping constant the ratio of any value 
of one to the corresponding value of the other, then one 



306 GENERAL MATHEMATICS 

is said to vary directly as the other or to be rtiwtly pro- 
portional to the other. Thus the number r is said to vary 

x 
directly as y if the ratio - remains constant, as x and y both 

x 
change or vary. The equation - = k expresses algebraically, 

i7 

and is equivalent to, the statement that JT varies directly 

X 

as y. The equation - = k is often written jc = ky. Show 
why this is correct. ^ 

EXERCISES 

Translate the following statements into equations of the 
form = /. : 

y 

1. The cost of 10 yd. of dress goods is directly propor- 
tional to the price per yard. 

Solution. I 'sing c for the total cost and j> for the price per yard, 

- = 10, or c = 10,,. 
P 

This illustrates direct variation, for the greater^the price per yard, 
the greater the total cost. 

2. The railroad fare within a certain state is 3 cents per 
mile. Write the equation, showing that the distance is directly 
proportional to the mileage. 

3. The weight of a mass of iron varies directly as the 
volume. 

4. If a body moves at a uniform rate, the distance varies 
directly as the time. 

5. The length (circumference) of a circle varies directly as 
the diameter. 

6. The distance d through which a body falls from rest 
varies directly as the square of the time t in which it falls. 
A body is observed to fall 400 ft. in 5 sec. What is the 
constant ratio of d to t 2 ? How far does a body fall in 2 sec. ? 



FUNCTION INTERPRETED GRAPHICALLY 307 
Solution. The equation for d and t is 

j 2 = k. Why.' 

In this problem *gg- = k ; 

hence k = 16. 

Substituting k = 16 and t' 2 = 2 2 in - = k, 



Solving, f/ = 64. 

This solution shows that & may be determined as soon as one 
value of t and the corresponding value of d is known. Once deter- 
mined, this value of k may be used in all problems of this type. 
Thus, in all falling-body problems k = 16 (approx.). 

7. How far does a body fall from rest in -3 sec. ? in 5 sec. ? 

8. If a: varies directly as y, and x = 40 when y = 8, find 
the value of x when y = 15. 

9. If ic varies directly as a-, and -w = 24 when x = 2, find 
w when x = 11. 

10. A stone fell from a building 576 ft. high. In how 1 many 
seconds did it reach the ground ? (Use the method of Ex. 6.) 

11. The speed of a falling body varies directly as the time. 
Write the equation for the speed V and the, time t. A body 
falling from rest moves at the rate of 180 ft. per second five 
seconds after it begins to fall. What will be the speed attained 
in nine seconds ?. 

12. The time t (in seconds) of oscillation of a pendulum 
varies directly as the square root of the length I ; that is, 

- = /,-. A pendulum 39.2 in. long makes one oscillation in 

v/ 

one second. Find the length of a pendulum which makes an 
oscillation in two seconds. 

13. The simple interest on an investment varies directly as 
the time. If the interest for 5 yr. on a sum of money is $150, 
what will be the interest for 6 yr. 4 mo. ? 



308 



GENERAL MATHEMATICS 



14. The weight of a sphere of a given material varies directly 
as the cube of its radius. Two spheres of the same material 
have radii of 3 in. and 2 in. respectively. If the first sphere 
weighs 6 lb., what is the weight of the second ? 

346. Graphing direct variation. Direct variation between 
two quantities may be represented graphically by means 
of a straight line. Turn back to Chapter XI, on the formu- 
las, and find three graphs illustrating direct variation. 



-20 



FIG. 217. GRAPH OF C = ird SHOWING DIRECT VARIATION 



An interesting example is furnished by graphing the 

C 
equation = TT (where TT = 3.14). This equation says that 

the circumference of a circle varies directly as its diameter. 

Complete the following table, and graph the results so 

as to obtain the graph in Fig. 217. Interpret the graph. 



d 





1 


2 


3 


4 


5 


6 


7 


8 


C 





3.1 


6.3 


9.4 













FUNCTION INTERPRETED GRAPHICALLY 309 

EXERCISES 

Graph the following examples of direct variation : 

1 . v = 16 1. (Velocity equals 16 times the number of seconds.) 

2. t = 5 /. (Turning tendency equals the weight of 5 times 
the lever arm.) 

3. A = 3b. (The area of a rectangle whose altitude is 3 
varies directly as the base.) 

347. Inverse variation. We shall now consider a new 
and interesting kind of variation. Suppose a gardener 
wishes to seed 64 sq. ft. of his garden in lettuce. If he 
makes it 16 ft. long, the width must be 4 ft. (Why?) If he 
makes it 32 ft. long, the width need be only 2 ft. (Why ?) 
How many possible shapes do you think the gardener 
might choose for his lettuce bed ? The following table 
will help you to answer this question if you remember 
that it has been decided that the area shall be 64 sq. ft. 



Length . . . 


40 


36 


32 


25 


20 


1(5 


8 


4 


> 


\ 


Width . . . 


1.6 


1.8 


2 


2.56 















The table shows that the length must vary so as to leave 
the area constant, and that because of this fact the 
greater the length, the smaller the width. The length is 
thus said to vary inversely as the width or to be inversely 
proportional to the width. Algebraically speaking, a number 
x varies inversely as y if the product xy remains constant 
as both x and y vary ; that is, if xy = k. The student may 

Jc k 

also find this equation written x = - or y = - 

y x 



310 GENERAL MATHEMATICS 

EXERCISES 

1. Express each of tin- following statements by means of 
equations : 

(a) The time needed to go a certain distance varies inversely 
as the rate of travel. 

(b) The heat of a stove varies inversely as the square of 
the distance from it. 

(c) Tke rate at which a boy goes to the corner drug store 
varies inversely as the time it takes him. 

18 

2. If .u = , show that w varies inversely with z. 

ir 

3. If y varies inversely as x, and x = 12 when y = 4, find 
the value of y when x = 2. 

Solution. By definition of inverse variation, 
xy = k. 

In the first case, x = 12, 

and y = 4. 

Therefore 1'J 4 = k, 

or k = 48. 

In the second case, x = 2. 

Therefore 2 y = 4S, since /. is constant. 

Then y = 24. 

4. If x varies inversely as ?/, and x = 12 when y = 13, find 
the value of x when y = 2. 

5. When gas in a cylinder is put under pressure, the vol- 
ume is reduced as the pressure is increased. The physicist 
shows us by experiment that the volume varies inversely as 
the pressure. If the volume of a gas is 14 cc. when the pres- 
sure is 9 lb., what is the volume under a pressure of 16 Ib. ? 

6. The number of men doing a piece of work varies inversely 
as the time. If 10 men can do a piece of work in 33 da., in 
how many days can 12 men do the same piece of work ? 



FUNCTION INTERPRETED GRAPHICALLY 311 



40 



30 



348. Graphing inverse variation. We shall now proceed 
to show how inverse variation may be represented graph- 
ically. Two cities are 48 mi. apart. Trains running at 
various rates carry the traffic between the two cities. 
Suppose we attempt 
to find out how 
long it will take a 
train which moves 
uniformly at the 
rate of 24 mi. per 
hour to make the 
trip, then 6 mi. per 
hour, 8 mi. per hour, 
etc. The following 
table contains some 
of the values by 
means of which the 
points in Fig. 218 
were plotted. The 
equation represent- 
ing the situation is 
48 = rt. When the 
points of the .table 
are plotted, it is clear that they do not lie on a straight 
line, as was the case in direct variation ; but if they are 
connected, the result is the curved line of Fig. 218. This 
line is one of two branches of a curve called a hyperbola. 



20 



Rate 



FIG. 218. GRAPH SHOWING INVERSE 
VARIATION 



r 


48 


32 


30 


24 


20 


16 


12 


8 


6 


4 

_ 


3 


2 


1 


t 


1 


1.6 


1.6 


2 





















312 GENERAL MATHEMATICS 

EXERCISES 

1. Determine from the graph in Fig. 218 the time it takes 
a train whose rate is 31 mi. per hour ; 20 mi. per hour ; 25 mi. 
per hour. 

2. Determine from the graph in Fig. 218 how fast a train 
runs which makes the trip in 1|- hr.; 2| hr.; 5^ hr.; 8^ hr. 

3. Graph the equation ^?y = 144 (see Ex. 5, Art. 347) and 
interpret the graph. 

4. The hyperbola is an interesting mathematical curve. See 
if you can help your class learn more about it by consulting 
other books. 

349. Joint variation. In the interest formula /= Prt, 
I depends for its value on P, r, and L A change in any 
one of these letters causes a corresponding change in the 
value of /. We express this by saying that the interest 
varies jointly as the principal, rate, and time. The alge- 
braic equation which defines joint variation is x = kyz. 

EXERCISES 

1. Turn to Chapter XI, on the formula, and find five illus- 
trations of joint variation. 

2. If z varies jointly as x and y, and if = 60 when x = 3 
and y = 2, find z when x = ^ and y = ^. 

*3. Write the following law as a formula: The safe load L 
of a horizontal beam supported at both ends varies jointly as 
the width w and the square of the depth d and inversely as the 
length I between the supports. 

*4. A beam 12 ft. long, 4 in. wide, and 8 in. deep when 
supported at both ends can bear safely a maximum load of 
1920 Ib. What would be the safe load for a beam of the same 
material 10 ft. long, 3 in. wide, and 6 in. thick ? 



FUNCTION INTERPRETED GRAPHICALLY 313 

SUMMARY 

350. This chapter has taught the meaning of the fol- 
lowing words and phrases : function, linear function, vari- 
able, dependent variable, independent variable, constant, 
direct variation, inverse variation, joint variation. 

351. A clear understanding of the dependence of one 
quantity upon another is very important in the everyday 
affairs of life. 

352. A linear function -may be treated algebraically, or 
geometrically by means of a graph. 

3/ 

353. The equation expressing direct variation is - = k. 

tj 

It is often written x = ky. Its graph is a straight line. 

354. The equation expressing direct variation is xy = &, 

k k 

x = -> or y = - Its graph is a curved line which is 

y x 

one branch of a hyperbola. 

355. The equation expressing joint variation is x = kzy. 



CHAPTER XIII 

SIMILARITY; CONSTRUCTION OF SIMILAR TRIANGLES; 
PROPORTION 

356. Construction of similar, triangles ; first method ; 
introductory exercises. The following exercises will help to 
form a basis for the work of this chapter. The student 
should study them carefully. 

EXERCISES 

1. On squared paper lay off a line segment AB of any con- 
venient length. At .1 construct, with the protractor, an angle 
of 32. At B construct an angle of 63 and produce the sides 
of the two angles so as to form a triangle. Call the vertex 
angle C. 

2. Compare the shape of the triangle ABC that you have 
drawn with the shape of those drawn by your classmates. 

3. What was done in Ex. 1 to insure that all members of 
the class should get triangles of the same shape ? 

4. With the protractor measure angle (.' in your figure. How 
else might you have determined its size ? 

5. Compare the size of angle C in your figure with angle C 
in the figures drawn by your classmates. 

6. Show that any angle C drawn by any member of the 
class ought to be equal to any other angle C drawn. 

7. Are the triangles drawn by the class of the same size ? 
Are any two necessarily of the same size '.' 

314 



CONSTRUCTION OF SIMILAR TRIANGLES 315 

357. Similar triangles. Triangles having the same shape 
are called similar triangles. Similar triangles are not neces- 
sarily of the same size. They may be constructed by making 
two angles of one equal to two angles of the other, as was 
done in Ex. 1, Art. 356. If two angles 6f one are equal 
to two angles of the other, it follows that the third angles 
are equal. (Why ?) The symbol for similarity is ~->. Thus 
AABC^AA'B'C" is read "triangle ABC is similar to 
triangle A'B'C'" The results of Art. 356 may be summed 
up in the following geometric theorem : If two angles of 
one triangle are equal respectively to two angles of another 
triangle, the triangles are similar. 

358. Second relation of parts in similar triangles. The 
student should be able to discover a second method of con- 
structing similar triangles if he studies and understands 
the following exercises. 

INTRODUCTORY EXERCISES 

1. In the triangle ABC drawn for Ex. 1, Art. 356, letter the 
side opposite angle C with a small letter c, the side opposite 
angle B with a small letter I, and the side opposite angle A with 
a small letter a. 

2. Measure the lengths of the sides a, l>, and c in Ex. 1 
to two decimal places. Find the ratio of a to I, of 1> to c, of 
a to c (in each, case to two decimal places). 

3. Compare your results in Ex. 2 with the results obtained 
by the other members of your class. What conclusion do you 
make with reference to the ratios of the sides '.' 

359. Construction of similar triangles ; second method. 
The results of Exs. 1-3, Art. 358, may be summarized as 
follows : In similar triangles the ratios of corresponding sides 
are equal. The work of Art. 358 suggests a second method 
for constructing similar triangles. 



3 It) GENERAL MATHEMATICS 

EXERCISES 

1. Draw a triangle. Draw a second triangle whose sides 
are respectively twice as long as the sides of the first 
triangle. 

2. Compare the triangles constructed in Ex. 1 as to shape. 
Are they similar ? Find the ratio* of the corresponding sides. 

3. Draw a triangle with sides three times as long as the 
corresponding sides of another triangle. Are they similar? 
Give reasons for your answer. How do the ratios of the 
corresponding sides compare ? 

4. Draw a triangle ABC. Bisect the lines AB, AC, and EC. 
Call the halves x', y\ and z'. Construct a second triangle, using 
the segments x', y', and z' as sides. Compare the two triangles 
as to shape. Are they similar ? What are the ratios of the 
corresponding sides ? 

The preceding exercises suggest the following theorem : 
Two triangles are similar if the ratios of the correspond- 
ing sides are equal. This gives us a second method of 
constructing similar triangles ; namely, by making the 
ratios of their corresponding sides equal. 

360. Construction of similar triangles ; third method. 
We shall study a third method of constructing similar 
triangles which is suggested by the following exercises : 

EXERCISES 

1. Construct a triangle with two sides 4.6cm. and 6.2cm., 
and with the protractor make the included angle 70. Con- 
struct a second triangle with two sides 9.2 cm. and 12.4 cm. 
and the included angle 70. Compare the triangles as to shape. 
What is the ratio of the corresponding sides ? Measure the 
corresponding angles. 



CONSTRUCTION OF SIMILAR TRIANGLES 317 



2. If convenient the class sholild divide itself into sections, 
,ne first section constructing a triangle with two sides and the 
included angle as follows : a = 12, b = 18, and C = 40; the 
second section taking a = 8, b = 12, and C = 40; and the third 
section taking a = 4, b = 6, and C 40. Compare the triangles 
drawn by the three sections as to shape. What is the ratio 
of the corresponding sides ? 

The preceding exercises support the geometric theorem: 
Two triangles are, similar if the ratio of two sides of one equals 
the ratio of two corresponding sides of the other, and the angles 
included between these sides are equal. This theorem suggests 
the third method of constructing similar triangles. 

361. Summary of constructions for similar triangles. Two 
triangles are similar 

1. If two angles of one triangle are constructed equal respec- 
tively to two angles of the second triangle. 

2. If the sides of the triangles are constructed so that the 
ratios of their corresponding sides are equal. 

3. If the triangles are constructed so that the ratio of two 
sides of one is equal to the ratio of two sides of the other and 
the angles included between 

these sides are equal. 

362. Similar right tri- 
angles. We shall now 
prove the following the- 
orem: The perpendicular 
to the hypotenuse from the 

vertex of a right triangle divides the triangle into two triangles 
that are similar to each other (see Fig. 219). 

Proof. x = /.x'. Why? 

Z. V = Z/. Why? 

.-. \ADC ^ AEDC. Why? 




D 



FIG. 219 



318 GENERAL MATHEMATICS 

EXERCISES 
(Exs. 1-4 refer to Fig. 219) 

1. Show that A.I IK' A ABC. 

2. Show also that A BCD ^ A.-lLv '. 

3. Translate the results of Exs. 1 and 2 into a geometric 
theorem. 

4. State a theorem expressing the results of this article. 

363. Similar polygons. In later work in mathematics 
we learn that similar polygons also have corresponding 
angles equal and that the ratios of the corresponding sides 
are equal. This rests on the fact that two similar polygons 
may be divided into sets of similar triangles by drawing 
corresponding diagonals us 
in Fig. 220. 

Similar figures are of 
frequent occurrence. The 
plans of construction work, FJG 22Q SIMILAK POLYGOXS 
drawings in shop, a sur- 
veyor's copy of a field triangle, blue prints, a photograph, 
enlarged and reduced pictures, are all examples. The rela- 
tion of the different parts in all the foregoing is shown by 
magnifying or reducing all parts to a definite scale. Thus, 
you may be able to determine by looking at a photograph 
of a man that he has large ears, although in the picture 
the actual measurement of either of his ears may be less 
than a centimeter. One can tell by looking at the plan 
of a house whether the windows are large or small, be- 
cause the relation is brought out by the fact that all parts 
are reduced to the same scale ; that is, the ratios of the 
corresponding parts are equal. See if you can find examples 
that will illustrate the last statement 





CONSTRUCTION OF SIMILAR TKI ANGLES 319 

Similar triangles may be regarded as copies of the same 
triangle magnified or minified to a scale, or both may be 
regarded as scale drawings of the same triangle to dif- 
ferent scales. We shall study the geometric relations more 
in detail in the^next chapter. 

364. Algebraic problems on similar figures. The fact 
that the ratios of the corresponding sides of similar poly- 
gons are equal furnishes us with an algebraic method of 
finding distances. 

EXERCISES 

1. In the similar triangles of Fig. 221, if a = 3 in., a' = 9 in., 
and b = 3 in., how long is // ? 

2. In the similar triangles of Fig. 222, if n = 6 -mm., 
a' = 8 mm., and I = 8 mm., how long 

is b 1 ? 

3. In the similar triangles of Fig. 223, 
if .a' =10.5 mm., b =12 mm., and // = a/\h 

15 mm., how long is a ? 

Flo 221 

4. The sides of a triangle are 16, 20, 

and 26. The shortest side of a similar 
triangle is 22. Find the other sides. 

5. The sides of a triangle are 2.3 cm., 
2.7 cm., and 3 cm. The corresponding 
sides of a similar triangle are a-, y, and 
12cm. Find the values* of x and y. 

6. Two rectangular boards are desired. 
One is to be 4 in. wide and 6 in. long, . 
the other is to be IS^in. long. How wide / \ 






should the second board be? FK , 223 

7. At a certain time of day a foot rule 

casts a shadow 10 in. long. How long is the shadow of a yard 
stick'at the same time? Draw a diagram and prove your work. 



820 



( ; K.\ KK A I. MATHEMATICS 



8. In Fig. 224 the pole, the length of its shadow, and 

the sun's rays passing over the top of the pole form a triangle. 

The shadow of the pole is measured, and is found to be 60 ft. 

long. At the same time the shadow of a vertical stick 2^ ft. 

high is measured, and is found to be 7|- ft. long. How may we 

determine the height of the pole 
without actually measuring it ? 






c 


\ 




\ 

\ 


h 


\ 








\ 




\ 



60' 

FIG. 22 1 



7.5' 
FIG. 225 



Solution. The stick? the shadow, and the sun's rays form a 

triangle similar to the first triangle (see Fig. 225). Why? 
If we let h denote the height of the pole, we get 

h 6 Why? 



Then 



2.5 7.5 

h = 20. 



z 






9. The shadow of a chimney is 85.2 ft. long. At the same 
time the shadow of a man 6 ft. 2 in. tall is 9 ft. 2 in. How high 
is the chimney ? @ 

10. Draw a triangle ABC on squared 
paper as in Fig. 226. Through a point D on 
AC draw line DE II AB. Measure the seg- A' 

ments CD, DA , CE, and EB to two decimal 

CD Cf 

places. Find the ratios and Howdo 

DA EB 

these ratios compare ? What does this show ? 

11. Draw a scalene triangle on squared 
paper, making the base coincide with one 
of the horizontal lines. Letter the triangles 

as in Fig. 227. Choose any line parallel to the base and letter 

it DE as in Fig. 227. Find the ratios and -7-- How do 

DA A B 

these ratios compare? State your conclusion as a theorem. 




FIG. 227 



CONSTR0C-TION OF SIMILAR TUI ANGLES 



n 




FIG. 221 



12. Suppose that in Fig. 227 Z>C =2.5 mi., DA = 7.5 mi., 
and CE = 9.ini., how long is EB ? 

13. In triangle ABC, Fig. 227, the line DE has been drawn 
parallel to the base AB. Prove that 

the small triangle CDE is similar to 
the large triangle ABC. 

14. IiiFig.228,if^U;=3,DE=5, 
and AB = 25.5, how long is BC ? 

15. Divide ; a line segment into 
two equal parts and show that your 
construction is correct. 

Construction. Let AH, Fig. 229, be 
the line segment. Draw A C through A, 
making any convenient angle with AB. 
On AC lay off AD and DE, each 
equal to 1 unit. Join E to B. Draw 
DF II EB. Show that ,1 F = FB. 

HINT, Use Ex. 10, p. 320. 



FIG. 229. How TO BISECT 
A LINE 



16-. Divide a line segment into 
two parts whose ratio is |. 

Construction. Let AB, Fig. 230, be 
the given segment. Draw A C, making 
any convenient angle with AB, as 
shown. On AC lay off AD = 2 units 
and DE = 3 units. Join E and B. 
Through D draw DF II EB. Show 
AF 2 



I)' 



Fir,. 230 



that 



FB 3 




17. Divide a line segment into 

3 1 2 a 
parts having the ratios-; ^; ^; y 

FIG. 231 

18. The distance across a swamp 

(Fig. 231) is to be found. A point C is located in the same 
line with A and B. At C and B lines CD and BE are drawn 




322 GENEKAL MATHEMATICS 

perpendicular to CB, and the line AD is drawn. The lengths 
of CB, DE, and EA are found to be 80 ft., 90 ft., and 250 ft 
respectively. Find the distance AB. 

19. Show that the distance A B across 
the swamp could also be found by meas- 
uring the lines shown in Fig. 232. 

365. Proportion. The preceding 
exercises dealing with similar triangles 
were solved by means of a special 

type of equation expressing the fact that two ratios in the 
geometric figure were equal. Thus in Fig. 233 the line AB 
is divided into two parts whose ratio is | (see the method 
of Ex. 16, Art. 364). In this construction it turns out that 
4F 2 

3- Wh .v'-' A. f. , B 

~2*" '' '' 

?. Why? Vs/ 



DE 3 

AF AD 
and =- Wh y ? Fio. 233 



Such an equation, which expresses equality of two ratios, 
is called a proportion. The line segments AF, FB, AD, and 
DE are said to be proportional, or in proportion. This 
means that AF divided by FB will always equal AD 
divided by DE. 

A proportion may thus be defined as an equation which 
expresses the equality of two fractions ; as, T 8 :r = f. Another 

ct c 
example of a proportion is - = - This may be read 

" a divided by b equals c divided by <?," or " a is to b as 
c is to rf," or " a over b equals c over d." Sometimes it is 
written a : b = c : d, but this form is not desirable. 



CONSTRUCTION OF SIMILAR TRIANGLES 323 

EXERCISE 

Is the statement f = f a proportion ? Give reasons for 
your answer. Is f = -^ a proportion ? Explain your answer. 

366. Means and extremes. The first and last terms in 
a proportion are called the extremes and the second and 

third terms the means. Thus, in the proportion - = a and 
d are the extremes and b and c the means. 

EXERCISES 

1. Compare the product of the means with the product of 
the extremes in the following proportions : 



What statement can you make concerning the products ? 

2. Make up several proportions and compare the product 
of the means with the product of the extremes. 

367. Theorem on the relation between the means and the 
extremes of a proportion. Exs. 1-2, Art. 366, illustrate a 
well-known law or theorem ; namely, that in a proportion 
the product of the means equals the product of the extremes. 

a. c 
If the given proportion is - = - then the law is algebraically 

stated thus : ad = be. 

The theorem may be proved as follows : 

a c 

Let - = - represent 
b d 

members by bd we get 



ft f* 

Let - = -. represent any proportion. Multiplying both 
b d 



abd _ cbd 

Reducing each fraction to lowest terms, 
ad be. 



324 GENERAL MATHEMATICS 

Since a proportion is a special kind of equation, there 

are special laws which often make a proportion easier 

to solve than other equations which are not proportions. 

The law given on page 323, Art. 307, is one of the many 

principles of proportion convenient to use. Thus, instead 

4 16 

of finding the L.C.D. in the equation - = and solving 

o . X 

in that way, we simply use the preceding law, and say 

4 x = 48. 
x = 16. 

The law is also a convenient test of proportionality, 
since it is usually simpler to find the products than to 
reduce the ratios to lowest terms. 

EXERCISES 

1. Test the following statements to see if they are 
proportions : 

3 _ 15 ' . l5 J 12/ 42 w _^ 21 m 

' 7 ~~ 35 ' ' 2^5 ~~ 1.4 ' 5 n ~ 2.5 n ' 

5 _ 8 2 a _ 5a 11.5 _ 7.7 

\~J ^ 77' v*) o ^~K " (*) o " o~o * 

<- 11 ox t .o x o.o Z.Z 

2. Find the values of the unknowns in the following pro- 
portions, and check by substituting in the original equations : 



(b) - .= ^ Solution. >/- + y - 20 = y z 9. 
66_l ' y -2 = -. 

} 10"5' 

f f -\\ y~ 12 3 11 + 5 11 -3 

< d >niT5' Check - irT3 = irri 

, 3+1 = 3^ + 2 i-> _ 8 

5 14 14" 7' 



CONSTRUCTION, ! (3F SIMILAR TRIANGLES 326 

] a - 13 = a - 14 : ^ ~T~ ~3 

I IL ' . /i\ x i O _ J- 

' /T _I- .^ ~~ _1_ 1 U/ . K ' 



3. If 5 and 3 are each added to a certain number, and 
1. and 2 are each subtracted from it, the four numbers thus 
obtained are in proportion. Find the number. 

4. Show how to divide a. board 54 in. long into two parts 
whose ratio is y^. 

5. What are the two parts of a line segment 10 cm. long 
if it is divided into two parts whose ratio is | ? 

6. The acute angles of a right triangle are as 2 is to 5 ; 
that is, their ratio is -|. Find the angles. 

7. If 10 be subtracted from one of two complementary 
angles and added to the other, the ratio of the two angles thus 
formed is ^. Find the angles. 

8. If 1| in. on a railroad map represents 80 mi., what 
distance is represented by 2| in.? 

9. Two books have the same shape. One is 4^ in. wide; 
and 7-g^ in. long. The other is 18 in. long ; how wide as it ? 

10. The records of two leading teams in the American 
League were Boston won 68, lost 32 ; Chicago won 64, lost 36. 
If the teams were scheduled to play each other ten more gauit-s. 
how many must Chicago have won to have been, tied with 
Boston ? 

11. If 1 cu. ft. of lime and 2 cm. ft. of sand are used in 
making 2.4 cu. ft. of mortar, how much of each is needed to 
make 96 cu. ft. of mortar ? 

368. Proportion involved in variation. Many problems 
in physics, chemistry, general science, domestic science, 
astronomy, and mathematics may be solved by either 
variation or proportion. In fact, the whole theory of 



826 GENERAL MATHEMATICS 

proportion is involved in our discussion of variation, but 
this fact is not always so obvious to a beginner. The fact 
that problems may be stated both in terms of variation and 
in terms of proportion makes it necessary for the student 
to recognize clearly the relation between variation and pro- 
portion. This relation will be illustrated in the following 
list of exercises. 

EXERCISES 

Solve by either variation or proportion : 

1. If 11 men can build a cement walk in 82 da., how long 
will it take 15 men to build it ? 

(a) Solution as a variation problem : 

mt = k. (The time it takes to build 
the walk varies inversely as 

the number of men.) 
Then 11 82 = k. 

Hence k = 902. 

Using this value of k in the second case, 

mt = 902 ; 
but m = 15. 

Whence 15 t = 902, 

and t = * T V = 60 T 2 S da. 

(b) Solution as a proportion problem. The number of men is not 
in the same ratio as the time necessary to build the walk, but in 
inverse ratio ; that is, 



This proportion means " the first group of men is to the second 
group of men as the time it takes the second group is to the time it 
takes the first group." 

Substituting the three known facts, 

II -!i 

15 7 82' 

Whence 15 d, = 902, 

and c/ = T* = 60 da. 



CONSTRUCTION OF SIMILAR TRIANGLES 327 

2. If 200 ft. of copper wire weighs 60 lb., what is the weight 
of 125 ft. of the same kind of wire ? 

3. The cost of wire fencing of a certain kind varies as the 
number of yards bought. If 12 rods cost $12.80, how much 
can be bought for $44.80? 

4. Two men are paid in proportion to the work they do. 
A can do in 24 da. the same work that it takes B 16 da. to do. 
Compare their wages. 

5. A farmer has a team of which one horse weighs 1200 lb. 
and the other 1500 lb. If" they pull in proportion to their 
weight, where must the farmer place the clevis on a four-foot 
doubletree so as to distribute the load according to the size 
of the horses ? 

369. Different arrangements of a proportion. The stu- 
dent will be interested in seeing in how many different 
forms a proportion may be arranged. This he may learn 
by solving the exercises that follow. 

EXERCISES 

1. Arrange the numbers 3, 6, 7, and 14 in as many propor- 
tions as possible. Do the same for the numbers 2, 5, 8, and 20. 

2. Can you write two ratios that will not be equal, using 
the numbers 2, 5, 8, and 20 as terms of these ratios ? 

3. How do you decide which arrangement constitutes a 
proportion ? 

The preceding exercises suggest that a proportion such 

as - = - may take four forms, as follows : 
o a 

(a) The given proportion 



328 GENERAL MATHEMATICS 

(b) The form obtained by alternating the means in (a): 

a b 
~c = d' 

(c) The form obtained by alternating the extremes in (a): 



(d) The form obtained by alternating both the means 
and extremes in (a) : 

d b 



The last form can be obtained simply by inverting the 
ratios- in (a). 

We know that the proportions given above are true, for by 
applying the test of proportionality we see that the product 
of the means equals the product of the- extremes in each 
case. Furthermore, any one of them could have been 
obtained by dividing the members of the equation ad = be 

by the proper number. Thus, to get = - we must divide 
both members of the equation ad = be by ab. Why ? 

. ad l<- 

Then = : 

ab ab 

from which T = - , or form (c). 

b a 

370. Theorem. The preceding discussion illustrates the 
use of the theorem which says that if the product of two 
numbers is equal to the product of two other munbers, either 
pair may be made the means and the other pair the extremes 
of a proportion. 



CONSTRUCTION OF SIMILAR TKIANGLES 329 

EXERCISES 

1. Start with the equation ad = be and obtain the forms 
a c a b d b 

7 = 3' - = 3 ' und - = - This completes the proof of the 
(> a c a c a 

theorem just stated. Why ? 

2. Write the four possible forms that can be obtained from 
the following products : 

(a) 5 21 = 7 15. (o) 3 a . 4 b = a 12 b. 

(b) 3 28 = 12 7. (d) 15 - 7 t = 105 /. . ( 



= . she, that =. 




FIG. 234 

4. If two equilateral polygons have the same number of 
sides, the corresponding sides are proportional (see Fig. 235). 

i n 

Proof. =1, Why? 

JJ o 

and 1 - 



TIV . A , r , .> 

Therefore ; = \v hy ? 

J5( /> C 

By alternating the means, 

,4 BC 



A'E' B'C' 
and so on for the other corresponding sides. 

371. Mean proportional. In the proportion -= -, b is 

called the mean proportional between a and c (note that 
b appears twice in the proportion). 



330 GENERAL MATHEMATICS 

EXERCISES 

1. What is a mean proportional between 4 and 9 ? 
HINT. Let x = the number. 

TU * X 

Ineii - = -. 

x y 

From which x- = 36. 

Then r = 6. 

2. Show that the value of b in the proportion - = - is given 

C 

by the equation b = Vac (read " & equals -f- or the square 
root of ac"~). 

3. "What is a mean proportional between 2 and 18 ? between 
10 and 40 ? between 2 and 800 ? 

4. Find a mean proportional between a? and i 2 ; between 
3? and y 3 . 

372. How to pick out corresponding sides of similar 
triangles. The similar triangles of Fig. 236 are -placed so 
that in certain cases a line is a side in each of two similar 
triangles. Thus, AC is a 
side of A ADC and also of 
the similar triangle ABC. 
This suggests that the 
line may occur twice in 
the proportion of the cor- 
responding sides. In this FlG 236 

way it is seen that the 
line becomes a mean proportional between the other two. 

This analysis can be checked only by actually writing 
the proportion of pairs of corresponding sides of similar 
triangles. In order to do this correctly the student must 
remember that he (nrrexpondinri sides of similar triangles 




CONSTRUCTION OF SIMILAR TRIANGLES 381 

are the sides which lie opposite equal angles. Hence, from 
the fact that A ADC > A ABC we may write the following 
proportion : 

AD (opposite Z z in A ADC) _ ^4 C (opposite Z 2: in A ADC) 
A C (opposite Z z' in A AB (7) ^(opposite Z.C in AACB) 

, AD AC 

1 hat is, = 

AC AB 

AC is thus seen to be the mean proportional between AD 
and AB. 

Show in a similar way that BC is a mean proportional 
between the hypotenuse AB and the adjacent segment BD ; 

BD BC . 

that is, show that - = - 
BC AB 

We may summarize the preceding exercises and dis- 
cussion by the theorem: In a right triangle either side 
including the right angle is a mean proportional between the 
hypotenuse and the adjacent segment of the hypotenuse made 
by a perpendicular from the vertex of the right angle to the 
hypotenuse. 

373. Theorem. If in a right triangle a perpendicular is 
drawn from the vertex of the right angle on the hypotenuse, 
the perpendicular is a mean proportional 
between the segments of the hypotenuse. 
The truth of the preceding theorem will 
be seen from the following: A D 

In AABC (Fig. 237) Z C is a right Fio. 237 

angle, and CD AB. |f = ff because A ADC - A CDS, 
and the corresponding sides are therefore in proportion. 




332 ; :...-. ..GENERAL MATHEMATICS-, ...;_ 

EXERCISES 

1. Write out the complete proof for the preceding theorem. 

2. Find the altitude drawn to the hypotenuse of a right 
triangle if it divides the hypotenuse into two segments whose 
lengths are 4 in. and 16 in. respectively. Find also each leg 
of the right triangle. 

*3. In a right triangle ABC (right-angled at C) a perpen- 
dicular is drawn from C to AB. If CD = 8, then A I) = 4. 
Find the length of AB. 

374. Construction of a mean proportional. The theorem 
of Art. ^73 on page 331 furnishes us with a method of con- 
structing a mean proportional between any | b [ 
two line segments, as will now be shown. 

In Fig. 238 we are given two line seg- ' - ' 
ments a and b. The problem is to con- 
struct a mean proportional (say x units long) between a and b. 

CL CC 

We know that the equation - = - will represent the 

x b 
situation, 

Construction. On a working line, as A K in Fig. 239, we lay off a 
from A to B and b from B to C: With AC as a diameter we 
construct a semicircle. At 
B we erect a perpendicular ^-- ~~~ 



^ 



B OK 



intersecting the circle at D. /' 

Then BD is the required 

mean proportional. ' /' 

I /' 
Proof. Connect A with D [^_ _ r JL 

and C with D. Then BD is l 

the required mean propor- FIG. 239. MEAN PROPORTIONAL 

tional between a and b pro- CONSTRUCTION 

vided we can show that Z-D 

is a right angle. (Why?) We shall proceed to show that ZD is a 

right angle by proving that if any point on a circle is connected with 

the ends of a diameter, the angle formed at that point is a right angle. 



CONSTRUCTION OF SIMILAR TRIANGLES 333 



In Fig. 240 we have a given circle constructed on the diameter 
AC and a point D connected with the ends of AC. We must 
show that ZZ) is a right angle. 

Connect D and 0. 

Then Z z = Z .s + Z .s', (1) 

because an exterior angle of a tri- 
angle is equal to the sum of the 
two nonadjacent interior angles; 



and 



Z y = Z t + Z I' 



(2) 



for the same reason. 

By adding (1) and (2), 




FIG. 240 



Since 

But 

and 

Therefore 

Then 



Z x + Z y = Z ,s- + Z / + Z t + Z ('. 
Z x + Z y - 180, 



Z < = Z <'. 

2 Z.v + 2 Z< = 180. 
Zs + Z< = 90. 
ZZ)--90. 



Why? 
Why ? 

Why? 

..Why? 

Why? 

Why? 



Then, if in Fig. 239 Z D = 90, the proportion - is true, 



and BD is a mean proportional between a and 



BD b 
Give reasons. 



EXERCISES 



1. Explain how a mean proportional between two given 
line segments may be constructed. 

2. Construct a mean proportional be- 
tween 9 and 16, 4 and 16, 4 and 9, 16 
and 25, 25 and 36. A 

3. In triangle ABC, Fig. 241, Z.C 

is a right angle, CDA.AB, AD = 2, and DB = 6. Find the 
lengths of AC and CB. 







334 GENERAL MATHEMATICS 

4. Find the mean proportional between the line segments 
m and n in Fig. 242. 

5. Measure in and n in Fig. 242 and the mean proportional 
constructed in Ex. 4. Square the value of m 

the mean proportional and see how the 

value compares with the product of m , n | 

and n - Fir.. 242 

*6. Construct a square equal in area to a 

given rectangle ; to a given parallelogram ; to a given triangle. 

a c 
375. Fourth proportional. In the proportion T = -^' d is 

\Jv 

called the fourth proportional to , 6, and c. There are 
two methods of finding the fourth proportional to three 
given numbers a, 5, and c. 

Algebraic method. Let x represent the value of the fourth 
proportional. 

TI, a c 

Then - = - 

ft x 

(by definition of a fourth proportional). 
Solving for x, ax = be. 
be 




x _ A c F G V 

a 

FIG. 243. How TO CONSTRUCT 

Geometric method. Take three given A FOURTH PROPORTIONAL 
lines, as a, b, and c in Fig. 243, and 

draw any convenient angle, as Z.BAC. On AB lay off AD = a, 
DE-l. On the other line A Clay oSAF=c. Draw DF. Then draw 
EG II DF as shown. Then FG is the required fourth proportional. 

See if you can show why the construction is correct. 

EXERCISES 

1. Check the construction above by measuring the four 

a c 

segments to see if = - 
o x 

2. Construct a fourth proportional to three given line seg- 
ments 2 cm., 3 cm., and 5 cm. long respectively. 



CONSTRUCTION OF SIMILAR TKIA^ULES 33o 









3. Show algebraically that the segment obtained in Ex. '1 
should be 1\ cm. long. 

4. Construct a fourth proportional to three segments 5 cm., 
6 cm., and 9 cm. long respectively. 

5. Check your work in Ex. 4 by an algebraic method. 

* 376. To find the quotient of two arithmetical numbers by 
a special method. To find || in per cent we need to solve the 

22 x 
equation ^ = w 

(Why?) This pro- 

portion suggests simi- 

lar triangles. If we 

take a horizontal line 

OM (Fig. 244) as 

a dividend line on 

squared paper, and 

ON perpendicular- to 

OM as a divisor line, 

then lay off OA on 

OM equal to 22 units, 

and at A erect a per- 

pendicular and mark 

off AB equal to 70 

units, we can solve 

our problem provided 

we draw another line DR 100 units above OM and parallel 

to it. Call DR the quotient line. 

Stretch a string fastened at so that it passes through 
J5, meeting the quotient line at C. 



Di 



de 



nd -Li 



-M- 



FIG. 244 



Then 



- or 



NOTE. The proof is left to the student. 
Therefore 22 is approximately 31% of 70. 



330 ( J KN K1J A L M ATH.EMATK 'S 

EXERCISES 

*1. Point out the similar triangles in the device for express- 
ing quotients "used in Fig. 244. Read the sides which are 
proportional. 

*2. A gardener planted 12 A. of potatoes, 8 A. of beans, 
13 A. of onions, 3 A. of celery, and 5 A. of cabbage. By means 
of the device used in Fig. 244 show the distribution of his 
garden in per cents. 

377. Verbal problems solved by proportion. We have 
said that many problems of science, the shop, and engineer- 
ing can be solved by proportion. We shall proceed to 
study how to solve some of these problems by using our 
knowledge of proportion. 

In the study of turning tendency, Art. 233, we recognized 
the following familiar principle of the balanced beam : The 
left weight times the left lever arm equals the right weight times 
the right lever arm. As a formula 

I~- 7-77 A F B 

tins may be written w l l l = u' 2 l z . i . . 



100 
Ib. 



60 
Ib. 



This principle is already familiar 
to all who have played with a 
seesaw. They discovered long ago Fi G 245 

that a teeter board will balance 

when" equal products are obtained by multiplying the 
weight of each person by his distance from the point of 
support (fulcrum). 

If, in Fig. 245, B weighs 60 Ib. and is 5 ft. from the ful- 
crum F, then A, who weighs 100 Ib., must be 3 ft. from 
the fulcrum. Thus, 60 5 = 100 3 is a special case of 
general law w l l 1 = w%l z . 

If we divide both members of the equation iv^^ = W 2 1 2 by 

7 nn 

w^ly we get - = - , which is in the form of a proportion. 
'2 w i 



CONSTRUCTION OF SIMILAR TRIANGLES 337 



The student will learn that in shop work many problems 
dealing with the lever or balanced beam may be solved by 
some form of the preceding proportion. 

BEAM PROBLEMS 

1. A lever (Fig. 246) 10 ft. long carries weights of 40 Ib. 
and 50 Ib. at its ends. Where should the fulcrum be placed 
so as to make the lever balance ? 

Solution. Let x = the number of feet from F to A , 

and 10 x = the number of feet from /' to B. 

Then 50 x = 40(10 -x). Why? A x F 10-x B 
50x = 400-40x. A A 



90 x = 400. 



50 

Ib. 



\40\ 
lib. 



FIG. 246 



2. A, who weighs 80 Ib., sits 6 ft. from the fulcrum. If B 
weighs 100 Ib., at what distance from the fulcrum should A 
sit in order to balance B ? 

3. A and B together weigh 220 Ib. They balance when A is 
5 ft. and B 6 ft. from the fujcrum. Find the weight of each. 

4. A and B are 4 ft. and 6 ft. respectively from the fulcrum. 
If B weighs 60 Ib., 

how much does A 
weigh ? 

5. How could 
you weigh yourself 
without a scale ? 

6. AB in Fig. 247 
is a crowbar 8| ft. 

long supported at F, \ ft. from A . A stone presses down at 
A with a force of 2400 Ib. How many pounds of force must 
be exerted by a man pressing down at />' to raise the stone ? 
(Disregard the weight of the crowbar.) 




FIG. 247 



GESEKAL MATHEMATK > 



7. In attempting to raise an automobile (Fig. 248) a man lifts 
with a force of 150 Ib. at the end of a lever 10 ft. long. The distance 
from the axle to F is 2 ft. 

What force is exerted up- 
ward on the axle as a result 
of the man's lifting ? 

8. Find ^ if 1 9 = 18 ft., 
>/- 2 = 62 Ib., and w 1 = 51 Ib. 

9. Find J a if ^ = 40 in., 
/- 2 =26 Ib., and ^=38 Ib. 




FIG. 248 



MIXTURE AND ALLOY PROBLEMS 

1. How much water must be added to 10 gal. of milk, 
testing \fo butter fat, to make it test 4% butter fat? 
Solution. Let x = the number of gallons of water added. 

Then x + 10 = the number of gallons of diluted milk, 

51 
and 10 = the amount of butter fat in the undiluted milk. 

I $ff ( x + 10) = the amount of butter fat in the diluted milk. 
Since the amount of butter fat remains constant, 

-$1 10= * (x 
100 

110 a: + 10 
200 



Why 'i 
Whv '{ 



25 
4 x + 40 = fjo. 

x = 3 1, the number of gallons of water to be added. 

2. A physician has a 25% mixture of listerine in water. How 
much water must he add to it to make it a 15 mixture ? 



Solution. Consider an arbitrary quantity of the mixture, say 
100 oz. 

Let x = the number of ounces of water added to every 

100 oz. of the mixture. 
Then 100 + x = the number of ounces in the new mixture. 



CONSTRUCTION OF SIMILAR TRIANGLES 339 

Since 25% of the original mixture is listerine, 

25 
= the per cent of listerine in the 

new mixture. 
And since 15% of the new mixture is to be listerine, 

25 = JL5_ 
100 + x 100 ' 

1500 +15 a: = 2500. 
15 x = 1000. 

Hence 66$ oz. of water must be added to every 100 oz. of the 
original mixture. 

3. How much water should be added to a bottle contain- 
ing 4oz. of the original mixture in Ex. 2 to make it a 15% 
mixture ? 

4. If a patent medicine contains 30% alcohol, how much 
of other ingredients must be added to 12 qt. of it so that 
the mixture shall contain only 20% alcohol ? 

5. How many quarts of water must be mixed with 30 qt. 
of alcohol 82 % pure to make a mixture 70 % pure ? 

6. "\Yhat per cent of evaporation must take place from 
a 5% solution of salt and water (of which 5% by weight 
is salt) to make the remaining portion of the mixture a 7% 
solution ? 

7. Two grades of coffee costing a dealer 25$ and 30$ per 
pound are to be mixed so that 50 Ib. of the mixture will be 
worth 28$ per pound. How many pounds of each kind of 
coffee must be used in the mixture ? 

8. In a mass of alloy for watch cases which contains 80 oz. 
there are 30 oz. of gold. How much copper must be added 
in order that in a case weighing 2 oz. there shall be -}>/. 
of gold ? 



340 GENERAL MATHEMATICS 

Solution. Let x = the number of ounces of copper to be added. 
Then 80 4- x = the number of ounces in the new alloy. 

- = the ratio between the whole mass of alloy 
30 

and the gold. 

y = the ratio of a sample of the new alloy to the 
gold in the sample. 

ri-,1 80 + X \ 

Then _ = _.. Why? 

80 + z = 120. Why? 



Hence 40 oz. of copper should be added. 

9. In an alloy of gold and silver weighing 80 oz. there are 
10 oz. of gold. How much silver should be added in order that 
10 oz. of the new alloy shall contain only ^ oz. of gold ? 

10. Gun metal is composed of tin and copper. An alloy of 
2050 Ib. of gun metal of a certain grade contains 1722 Ib. of 
copper. How much tin must be added so that 1050 Ib. of the 
gun metal may contain 861 Ib. of copper ? 

*378. Specific-gravity problems. A cubic foot of glass 
weighs 2.89 times as much as a cubic foot of water (a 
cubic foot of water weighs 62.4 Ib.). The number 2.89 is 
called the specific gravity of glass. In general, the specific 
gravity of a substance is defined as the ratio of the weight 
of a given volume of the substance to the weight of an equal 
volume of water at 4 centigrade. What would it mean, 
therefore, to say that the specific gravity of 14-karat gold 
is 14.88 ? A cubic centimeter of distilled water at 4 cen- 
tigrade weighs just 1 gm. Since the specific gravity of 
14-karat gold is 14.88, one cubic centimeter of gold weighs 
14.88 gm., 2 cc. weighs 29.76 gm., etc. In short, the weight 
of an object in grams equals the product of its volume in 
centimeters times its specific gravity. 



CONSTRUCTION OF SIMILAR TRIANGLES 341 

EXERCISES 

1. How many cubic centimeters of distilled water (specific 
gravity equal to 1) must be mixed with 400 cc. of alcohol 
(specific gravity equal to 0.79) so that the specific gravity of 
the mixture shall be 0.9? 

HINT. Find the weight of the two parts and set the sum equal 
to the weight of the mixture. 

2. Would you accept half a cubic foot of gold on the con- 
dition that you carry it to the bank ? Explain your answer. 

3. Brass is made of copper and zinc. Its specific gravity is 8.5. 
How many cubic centimeters of copper (specific gravity 8.9) must 
be used with 100 cc. of zinc (specific gravity 7.15) to make brass ? 

4. What is the specific gravity of a steel sphere of radius 
1 cm. and weight 32.7 gm. ? 

379. Proportionality of areas. The geometric exercises 
to be given in this article are important. The student 
should study them carefully, prove them, and try to 

illustrate each. 

EXERCISES 

1. Prove that the areas of two rectangles are to each other 
as the products of their correspond- 
ing dimensions. 

Proof. Denote the areas of the rec- 
tangles by #j and 7? 2 , as in Fig. 249, 

and their dimensions as shown. ' 

R l = a l b l . Why? 

R z = a 2 b 2 . Why? 

Therefore 5l = 2LI. Why? 

R 2 a A FIG. 249 

It is important to note that the last proportion is obtained by 
dividing the members of the first equation by those of the second. 





842 GENERAL MATHEMATICS 

2. If two rectangles (Fig. 250) have equal bases, they are 
to each other as their altitudes. (Follow the method of Ex. 1.) 

3. If two rectangles have equal alti- 
tudes, they are to each other as their bases. 

4. The area of a rectangle is 48 sq. ft. 
and the base is 11 yd. What is the area of 
a rectangle having the same altitude and 
a base equal to 27.5 yd. ? 

5. .Prove that the areas of two parallelo- 
grams are to each other as the products of 
their bases and altitudes. 

6. The areas of two triangles are to 

each other as the products of their bases and altitudes. 

7. The areas of two parallelograms having equal bases are 
to each other as their altitudes ; the areas of two parallelograms 
having equal altitudes are to each other as their bases. 

8. The areas of two triangles having equal bases are to 
each other as their altitudes ; the areas of two triangles having 
equal altitudes are to each other as their bases. 

9. Prove that triangles having equal bases and equal alti- 
tudes are equal. 

*10. Construct the following by means of Ex. 9: a right tri- 
angle, an isosceles triangle, an obtuse-angled triangle, each equal 
to a given triangle. 

SUMMARY 

380. This chapter has taught the meaning of the fol- 
lowing words and phrases : similar triangles, similar 
polygons, proportion, means, extremes, fulcrum, mean 
proportional, fourth proportional, alloy, specific gravity. 

381. Polygons that have the same shape are similar. 

382. In similar triangles the corresponding angles are 
equal and the corresponding sides are in proportion. 



CONSTRUCTION OF SIMILAR TRIANGLES 343 

383. Two similar triangles may be constructed by 

1. Making two angles of one equal to two. angles of 
the other. 

2. Making the ratios of corresponding sides equal. 

3. Making the ratio of two sides of one equal to the 
ratio of two sides of the other, and the angles included 
between these sides equal. 

384. A proportion expresses the equality of two ratios. 

385. A convenient test of proportionality is the theorem 
that says the product of the means equals the product of 
the extremes. 

386. If ad = be, we may write the following four 
proportions : 

- a c a b d c ... d b 



387. The fact that the ratios of corresponding sides of 
similar polygons are equal furnishes us with an algebraic 
method of finding distances. 

388. Inaccessible distances out of doors may often be 
determined by means of a proportion. 

389. Beam problems and mixture, alloy, and specific- 
gravity problems may be solved by equations which take 
the form of proportions. 

390. If a line is drawn parallel to the base of a triangle, 
the triangle cut off is similar to the given triangle, and the 
corresponding sides are in proportion. 

391. The following important theorems about the area 
of two parallelograms have been proved : 

1. The areas of two parallelograms are to each other 
as the product of their bases and altitudes. 



344 GENERAL MATHEMATICS 

2. The areas of two parallelograms having equal bases 
are to each other as their altitudes, and the areas of two 
parallelograms having equal altitudes are to each other as 
their bases. 

392. Three theorems similar to those in Art. 391 were 
proved for the areas of rectangles and triangles. 

393. If in a right triangle a line is drawn from the 
vertex of the right angle perpendicular to the hypotenuse, 

1. The triangle is divided into two similar triangles 
which are each also similar to the given triangle. 

2. The perpendicular is a mean proportional between 
the two segments of the hypotenuse. 

3. Either side about the right angle is a mean propor- 
tional between the whole hypotenuse and the adjacent 
segment. 

394. The following constructions have been taught: 

1. How to construct a mean proportional. 

2. How to construct a square equal to a given rec- 
tangle, parallelogram, or triangle. 

3. How to construct a right triangle or an isosceles 
triangle equal to a given scalene triangle. 

4. How to construct a fourth proportional. 

5. How to divide a line segment into two parts which 
have a given ratio. 



CHAPTER XIV 

INDIRECT MEASUREMENT; SCALE DRAWINGS; 
TRIGONOMETRY 

395. Scale drawings. Up to this point we have made 
several uses of scale drawings. In Chapter X the relation of 




FIG. 251. SCALK DRAWING OF A LIBRARY TABLE 
(Courtesy of Industrial Arts Magazine) 

quantities was shown by line segments whose proportional 
lengths represented the relative size of the quantities. 



345 



34l> GEXE11AL MATHEMATICS 

In another form of graphic work, scale drawings have 
helped us to understand the meaning of functions, equa- 
tions, and formulas. In addition to the foregoing, scale 
drawings are probably familiar to the student in the form 
of shop drawings, geography maps, blue prints, maps in 
railroad guides, and architects' plans. 

The shop drawing in Fig. 251 illustrates a use of a scale 
drawing, which we shall now study in some detail. 

The figure shows that a scale drawing gives us an accu- 
rate picture of the real object by presenting all the parts 
in the same order of arrangement and showing the relative 
sizes graphically by means of proportional line segments. 
Obviously this fact rests on the principle of similarity, and 
the ratio between any two line segments in the plan equals 
the ratio between the lengths of the two corresponding 
parts of the library table (Fig. 252). 

By means of the scale drawing we are able to deter- 
mine the dimensions of parts of the table even though 
they are not given on the plan. In fact, in the case of 
architects' and surveyors' scale drawings we are able to 
measure lines which in the real object are inaccessible. 

This last procedure illustrates precisely the use which 
we want to make of scale drawings in this chapter. In 
many cases we shall want to measure distances that can- 
not be measured directly with steel tape or other surveying 
devices; for example, (1) the heights of towers, buildings, 
or trees ; (2) the width of ponds, lakes, or rivers ; (3) the 
length of boundary lines passing through houses, barns, or 
other obstructions. 

We can usually determine such distances by following 
the method set forth in the following outline: 

1. Measure enough actual lines and angles in the real 
object so that a scale drawing of the object can be made. 



TRIGONOMETRY 



347 



2. Draw the figure to scale, preferably on squared paper. 

3. Measure carefully with the compasses and squared 
paper the lines in the figure which represent the inaccess- 
ible lines of the actual object that is being considered. 




FIG. 252. THE FINISHED LIBRARY TABLE 
(Courtesy of Industrial Arts Magazine) 

4. Translate the measurements obtained in (3) into the 
units used in measuring the lines of the actual figure. 

EXERCISES 

1. A man walks from his home around a swamp (Fig. 253). 
He starts from his home at A, walks 0.95 mi. north, then 1.2 mi. 

east, then 0.35 mi. south. How far from B i 2ml 

home is he ? 

Solution. Let 2 cm. represent 1 mi. Make a 
drawing on squared paper of the distances as 
shown in Fig. 253. Then on the squared paper 
a side of every small square represents 0.1 mi. 
(Why ?) The required distance is the number 
of miles represented by the segment A D, which 
is 13.9 small units long. Hence AD represents 1.39 mi. Why? 




348 GENERAL MATHEMATICS 

2. Show how the four steps given in the outline of Art. 395 
are followed in the solution of the preceding problem. 

3. A man starting at a point S walks 48 yd. north and then 
56 yd. east. Find the direct distance from the stopping-point 
to the starting-point. (Let 1 cm. = 10 yd.) 

4. A man walks 92 yd. south, then 154 yd. east, and then 
132 yd. north. How far is he from the starting-point ? (Use 
1 cm. for every 12 yd.) 

5. Two men start from the same point. One walks 15 mi. 
west, then 9 mi. north ; the other walks 12 mi. south, then 
16 mi. east. How far apart are they ? 

6. Draw to scale a plan of your desk top and find the 
distance diagonally across. (Use the scale 1 cm. = 1 ft.) 

7. A baseball diamond is a square whose side is 90 ft. 
By means of a scale drawing, find the length of a throw from 
" home plate " to " second base." 

8. The broken line ABC 
(Fig. 254) represents a coun- 
try road. Find out how much 
nearer it would be to walk diag- 
onally across country from A to A 3 ' 3 mi> B 

C than it is to follow the road. FlG - 254 

9. A roadbed is said to have a 6% grade when the level 
of the road rises 6 ft. in 100 ft. measured horizontally. ' Draw 
to scale a roadbed 520 yd. long which has a 6% grade. 

10. The sides of a triangular chicken lot are 20 ft., 16 ft., 
and 18 ft. respectively. Make a scale drawing of this lot on 
squared paper and estimate the area by counting the small 
squares and approximating the remaining area. 

11. In a map drawn to the scale of 1 to 200,000 what lengths 
will represent the boundaries of a rectangular-shaped county 
40 mi. long and 20 mi. wide ? Give the answer to the nearest 
hundredth of an inch. 




TRIGONOMETRY 



349 



12. A railroad surveyor wishes to measure across the swamp 
at A B represented in Fig. 255. He measures the distance from 
a tree A to a stone C and finds it to be 110 yd. 
Find the distance across the swamp if the 
angle at C is 70 and if BC = 100 yd.- 

NOTE. The lines BC and AC are meas- 
ured by means of a steel tape (Fig. 256) or a 
surveyor's chain (Fig. 257). The angle at C is 
measured by means of a transit (Fig. 43). Chaining pins (Fig. 258) 
are used by surveyors to mark the end-points of the chain or tape. 







FIG. 256. STEEL 
TAPE 



FIG. 257. SURVEYOR'S 
CHAIN 



FIG. 258. CHAIN- 
ING PINS 



* 13. If available examine a steel tape, chain, and the pins used 

by surveyors and report to class on the length, graduations, etc. 

14. In Fig. 259, S represents a water-pumping station in 

Lake Michigan. A and B represent two Chicago buildings on 





the lake shore. Reproduce the measurements to scale and 
determine the distance of S from each of the two buildings. 

15. In Fig. 260" a swimming course AB across a small lake 
is represented. Find A B by means of a scale drawing. 



350 GENERAL MATHEMATICS 

16. A triangular lot has these dimensions: .4.6 = 20 yd.; 
BC 40 yd.; A C = 30 yd. Make a scale drawing of the lot on 
squared paper and determine its area. (Since the formula for 

the area of a triangle A = calls for an altitude, the student 

L 

will draw one from A to BC and then apply the formula.) 

*17. In order to measure the distance between two pump- 
ing stations A and B in Lake Michigan a base line C/>=18.8 
chains long (1 chain = 66 ft.) was measured along the shore. 
The following angles were then measured: Z. A CD = 132 ; 
Z.BCD= 50; Z.CDA = 46; ZCZ> = 125. Draw the figure 
to scale and find the distance from A to B in feet. 

*18. Two streets intersect at an angle of 80. The corner 
lot has frontages of 200 ft. and 230 ft. on the two streets, and 
the remaining two boundary lines of the lot are perpendicular 
to the two streets. What is the length of these two boundary 
lines ? What is the area of the lot ? 

HINT. Construct the two perpendiculars with compasses. Then 
draw a diagonal so as to form two triangles and construct their 
two altitudes as in Ex- 16, above. 

*19. A class in surveying wishes 
to determine the height of a smoke- 
stack as shown in Fig. 261. The 
transit is placed at B, and the angle 
ij is found to be 62: then at A, 
and the angle x is found to l>e 32. 

Line AB is 48 ft. long and is measured along level ground. 
The transit rests on a tripod 3^ ft. high. Find the height 
of the chimney. 

396. Angle of elevation. The angles x and y which 
were measured in Ex. 19 are called angles of elevation. 
The angle KAH in Fig. 262 shows what is meant by an 
angle of .elevation. To. find .the angle of elevation, the 



^x* . 


A 

...-' / 
/ 
B -. y 


I 




?B0n 




TRIGONOMETRY 

transit is placed at A as in Fig. 262. The telescope 

of the transit is first pointed horizontally toward the 

smokestack. The farther end is then ^ 

raised until the top of the chimney 

K is in the line of sight. The angle 

KAH, through which the telescope A < JL tal Une \ { 

turns, is the angle of elevation of K 

. . , FIG. 262. ANGLE OK 

irom A, the pouit or observation. ELEVATION 



EXERCISES 

By means of scale drawings, compasses, and protractor 
solve the following exercises : 

1. When the angle of elevation of the sun is 20 a building 
casts a shadow 82 ft. long on level ground. Find the height of 
the building. 

2. Find the angle of elevation of the sun when a church 
spire 80 ft. high casts a shadow 120 ft. long. 

3. A roof slopes 1 in. per horizontal foot. What angle does 
the roof make with the horizontal ? 

4. A light on a certain steamer is known to be 30 ft. above 
the water. An observer on the shore whose instrument is 
4 ft. above the water finds the angle of elevation of this 
light to be 6. What is the distance from the observer to 
the steamer ? 

5. What angle does a mountain slope make with a horizontal 
plane if it rises 150 ft. in a horizontal distance of one tenth 
of a mile ? 

6. The cable of a captive balloon is 620 ft. long. Assuming 
the cable to be straight, how high is the balloon when^all 
the cable is out if, owing to the wind, the cable makes an angle 
of 20 with the level ground (that is, the angle of elevation 
is 20) ? 



352 GENERAL MATHEMATICS 

7. On the top of a building is a flagpole. At a point A on 
level grour/4 70 ft. from the building the angle of elevation of 
the top of the^lagpole is 42. At the same point, A, the angle 
of elevation of the top of the building is 32. Find the height 
of the flagpole. How high is the building ? 

397. Angle of depression. A telescope at M in the top 
of a lighthouse (Fig. 263) is pointed horizontally (zero 
reading), and then the farther end is lowered (depressed) 
until the telescope points to a boat at B. The angle HMB, 
through which the telescope turns, is the angle of depression 
of the boat from the point M. In Fig. 26 3, Z HMB = ^MBC. 
Why is this true ? 

EXERCISES 

1. If the height of the lighthouse (Fig. 263) is 220 ft. 
above water, and the angle of depression of the boat, as seen 

from M. is 40, what is the distance 

. . . H horizontal line M 

of the boat from R if R C is known 

to be 40ft.? 

2. A boat passes a tower on which 
is a searchlight 220 ft. above sea level. 
Find the angle through which the B 

beam of light must be depressed from FlG - 263 - ANGLE OF 

,i , , i DEPRESSION 

the horizontal so that it may shine 

directly on the boat when it is 300 ft. from the base of 
the tower. 

3. How far is the boat from the base of the tower if the 
angle of depression" is 51 ? 30 ? Xote that the height of the 
lighthouse is known, and that the distance of a boat out at sea 
depends on the size of the angle ; that is, the distance is a 
function of the angle. In other words, the lighthouse keeper 
needs only to know the angle of depression to determine the 
distance of a boat at sea. 




TRIGONOMETRY 



353 



4. From the top of a mountain 2500 ft. above the floor of 
the valley the angles of depression of two barns in the level 
valley beneath, both of which were due east of the observer, 
were found to be 27 and 56. What is the horizontal distance 
between the two barns ? 

5. From the top of a hill the angles of depression of two 
consecutive milestones on a straight level road, running due 
south from the observer, were found to be 23 and 47. How 
high is the hill ? 

398. Reading angles in the horizontal plane ; bearing of 
a line. The special terms used in stating problems in sur- 
veying and navigation make the problems seem unfamiliar, 
although the mathematics is frequently not more difficult 



N 



N 



N 



N 



w- 



-EW- 



-E 



S s s s 

FIG. 264. ILLUSTRATING THE BEARING OF A LINK 

than iii the exercises given in Art. 397. Thus the acute 
angle which a line makes with the north-south line is the 
bearing of the line. 

The bearings of the lines indicated by the arrows in 
Fig. 264 are read as follows : 75 east of north, 46 west 
of north, 20 east of south, and 30 west of south. These 
are written more briefly as follows: N. 75 E., N.46W., 
S.20E., and S.30W. 



354 



GENEKAL MATHEMATICS 



EXERCISES 

1. Read the bearings of the arrow lines in Fig. 265. 

2. W T ith a ruler and protractor draw lines having the fol- 
lowing bearings : 

(a) 26 east of south. (d) 37^ west of south. 

(b) 39 east of north. (e) 33 west of north. 

(c) 40 west of north. (f) 3 east of south. 

3. Write in abbreviated form the bearings of the lines in Ex. 2. 

N N N N ,V 

*f 



W- V -E W- E W-- 



-E W^- E W- -h^-- 



s s s 

FIG. 265 

399. Bearing of a point. The bearing of a point B 
(Fig. 266) from a point is the bearing of the line OB 
with reference to the north-south line through 0. 

EXERCISES 

1. In Fig. 266 read the bear- 
ing of 

(a) A from O. (d) from B. 

(b) from A. (e) C from O. 

(c) B from O. (f) from C. 

2. Point A is 6.4 mi. east 
and 9.8 mi. north of B. Find 
the distance from A to B. What 
angle does AB make with the 
north-south line through B? 
What is the bearing of B 
from A? of A from B? 

3. Sketch the figure for Ex. 2 

and show why the angles appearing as results for Ex. 2 are equal. 




TRIGONOMETRY 355 

4. The bearing of a fort B from A, both on the seacoast, is 
N. 55 W. An enemy's vessel at anchor off the coast is observed 
from A to bear northwest; from B, northeast. The forts are 
known to be 8 mi. apart. Find the distance from each fort to 
the vessel. 

400. The limitations of scale drawings. By this time the 
student probably appreciates the fact that a scale drawing 
has its limitations. He would probably not agree to buy 
a triangular down-town lot whose altitude and area had 
been determined by a scale drawing. If a millimeter on 
the squared paper represents 0.1 of a mile, a slight slip 
of the pencil or compasses means disaster to accuracy. 

Scale drawing is used extensively by the surveyor and 
engineer in the following ways: (1) as a method of esti- 
mating probable results; (2) as a help to clear thinking 
about the relations of lines and angles involved in a geomet- 
ric drawing ; (3) as a valuable check on results obtained by 
more powerful methods. But as a matter of fact we need a 
more refined method to determine lines and angles where 
a high degree of accuracy is desirable. We shall now pro- 
ceed to consider a far more efficient method of determining 
such lines and angles. Most students will find the method 
fascinating, because the solution is simple and the results 
obtained are as accurate as the lines and angles which are 
directly measured. 

TRIGONOMETRY 

401. Similar right triangles. A few exercises on similar 
right triangles will help the student to understand the 
new and more accurate method of determining lines and 
angles. This method may be used independent of scale 
drawings, is shorter in most cases, and lays the foundation 
for future mathematical work. 



GENERAL MATHEMATICS 




a 



b O 

<;. 207 



EXERCISES 

1. With the protractor construct a right-angled triangle 
having an angle of 37. Letter the figure as suggested in 
Fig. 267. Measure the lines a, b, and c. Let 

2 cm. represent 1 unit. Find the value of the 

ratios - - > and to two decimal places. 
c c l> 

2. Compare your result with other mem- 
bers of the class. Did all members of the class 
use the same length for the bases ? Are any 

two of the triangles drawn necessarily of the same size ? Show 

why the result obtained for 

c 

should be the same number as 
the results of your classmates. 

3. Prove that if two right 
triangles have an acute angle 

of one equal to an acute angle of the other, the ratios of their corre- 
sponding sides are equal. Write two proportions. (Use Fig. 268.) 

4. Could you draw a right triangle with angle A =37 in 

which - does not equal approximately 0.60, or ~ ? Prove. 

HINT. The fact that - = - means that in every right triangle the 
side opposite a 37 angle is approximately as long as the hypotenuse. 

5. A balloon B (Fig. 269) is fastened by a cable 200 ft. long. 
Owing to the wind the cable is held practically straight and 
makes a 37 angle with the horizontal. How high is the balloon ? 

Solution. This triangle is similar to every tri- B 

angle drawn by the class in Ex. 1. Prove. 




Therefore 



= O.GO. 
200 

a = 120 ft. 




\37 



90; 



Solving, 

Note that the solution is exceedingly simple 
(only two equations) and that the accuracy of 
the result does not now depend upon the accuracy of Fig. 269. 



b 
FIG. 269 




TRIGONOMETRY 357 

402. Sine of an angle. The ratio - (Fig. 270) is called the 
sine of the angle A. The abbreviation for "sine" is"sin." This 

definition may be written sin A - Thus, 

c 

sin 37 = = O.GO (approx.). Do you think ' 

we would have obtained the same value for - 

c A b C 

if in Ex. 1 we had made the angle 47 ? FHJ. 270 

EXERCISES 

1. Find the sine of 20, using the definition given in Art. 402. 
HINT. As in Ex. 1, Art. 401, construct the triangle, measure a 

and c, and find the value of - to two decimal places. 

c 

2. Find the sine of each of the following angles : 10, 15, 
2u, 32, 47, 68, 87. Compare each result with the results 
of your classmates. 

403. Table of sines. The preceding exercises show that 
the sine of the angle changes with the angle ; that is, 
sin 68 is not equal to sin 37. By taking a large sheet of 
graphic paper and a very large unit we could get a fairly 
good table, but it would be too much trouble to do 'this 
for every problem. Such a table has been very carefully 
calculated for you in the first column of the table in 
Art. 410. 

EXERCISE 

Turn to the table in Art. 410 and .see how efficient yo\i 
have been by comparing your results for Ex. 2, Art. 402, with 
the table. 

404. Cosine of an angle. The exercises given in this 
article will introduce another trigonometric ratio. 



358 GEJSEKAh MATHEMATICS 

INTRODUCTORY EXERCISES 

1. Construct a right-angled triangle with angle .1 (see 
Fig. 270) equal to 43. Measure b and c. Find the quotient 

-.to two decimal places. Compare the results with those of 
the other members of the class. 

2. Show that all results ought to agree to two decimal 
places. The ratio - (Fig. 270) is called the cosine of the angle A . 

C 

The abbreviation for " cosine " is " cos." Thus, cos 43 = 0.73 
(approx.). This means that in an}- right-angled triangle the 
side adjacent to the angle 43 is about -j 7 ^- as long as the 
hypotenuse. 

3. Find the cosine of 5, 18, 25. 35, 47, 65, 87. 

4. Compare the results for Ex. 3 with the table of cosines 
in Art. 410. 

405. Tangent of an angle. We shall now introduce a 
third important ratio connected with similar right triangles. 
Historically the tangent ratio came first. We shall have 
occasion to learn more about it. 

INTRODUCTORY EXERCISES 

1. In Fig. 270, what is the value of - ? Compare your result 
with the results obtained by other members of the class. 

2. Show that all the results obtained for - in Ex. 1 should 

b 
agree. 

The ratio - is called the tangent of angle. A. In speak- 
ing of the tangent of 43 we mean that the side opposite 
angle A is y 9 ^ (approx.) of the length of the side adjacent. 
The abbreviation for " tangent" is "tan." Thus, tan 45 = 1. 



TRIGONOMETRY 359 

EXERCISES 

1. Find the tangent of 11, 36, 45, 57, 82. 

2. Compare the results of Ex. 1 with the table of tangents 
in Art. 410. 

406. Trigonometric ratios. Solving a triangle. The ratios 

a b , a 

-> -> and - are called trigonometric ratios. We shall now 

proceed to show that the use of these ratios greatly 

simplifies the solution of many problems in- 

volving indirect measurements. By their use 

any part of a right triangle can be found if 

any two parts (not both angles) besides the 

right angle are given. This process is called 




solving the triangle. 

FIG. 271 

407. Summary of definitions. The follow- 

ing outline will be found convenient to the student in 
helping him to remember the definitions (see Fig. 271): 

a side opposite 
1. sm A=- - C- 

c lii/potennse 

b side adjacent 

'2. cos A = - = 

i' hypotenuse 

a side opposite 

3. tan A = - = - 

o side adjacent 

408. Trigonometric ratios clear examples of the function 
idea. Either by your own crude efforts at building a table 
or by a study of the table of ratios given, it is easy to see 
that the value of the ratio changes as the angle changes: 
that is, a trigonometric ratio depends for its value upon 



360 GENERAL MATHEMATICS 

the size of the angle. Hence the ratios furnish us with 
one more clear example of the function idea. We may 
therefore refer to them as trigonometric functions. 

HISTORICAL NOTE. Trigonometric ratios are suggested even 
in the Ahmes Papyrus (c. 1700 B.C. ?), which, as has been stated, 
may itself be a copy of some other collection written before the 
time of Moses. In dealing with pyramids Ahmes makes use of 
one ratio that may possibly correspond roughly to our cosine and 
tangent. 

The first to make any noteworthy progress in the development of 
trigonometry was Hipparchus, a Greek, who lived about 150 n.r. 
He studied at Alexandria, and later retired to the island of Rhodes, 
where he did his principal work. He was able to calculate the length 
of a year to within six minutes. 

The Hindus contributed to the early development of the science, 
from about A.D. 500, and the Arabs added materially to their work 
from about A.D. 800 to A.D. 1000 

Regiomontanus (or Johann Miiller, 1436-1476), a German, freed 
the subject from its direct astronomical connection and made it an 
independent science. 

In the sixteenth century the subject developed slowly, but in the 
seventeenth century it made a very decided advance, due to the 
invention of logarithms, mentioned later, and to the great improve- 
ment of algebraic symbolism which made it possible to write trigo- 
nometric formulas in a simple manner. Trigonometry in the form 
that we know it may be said to have been fully developed, except 
for slight changes in symbols, in the seventeenth century. 

409. Table of trigonometric ratios of angles from 1 to 89. 
The student should now become familiar with the table 
on the following page. The ratios are in most cases only 
approximate, but are accurate enough for all ordinary 
work. 

410. The use of a trigonometry table. The problems 
beginning on page 362 are intended to furnish the student 
practice in the use of the table. 



TRIGONOMETRY 



361 



Angle 


Sine 

(oPP.\ 
\hvp.) 


Cosine 
tad\ 
\hupj 


Tangent 
(opp.\ 
\adj.) 


Angle 


Sine 
(oM>.\ 
\hvP.J 


Cosine 
/adJA 
\hyp.) 


Tangent 

(opp.\ 
\adj.) 




1 
2 
3 
4 


.000 
.017 
.035 
.052 
.070 


1.000 
1.000 
.999 
.999 
.998 


.000 
.017 
.035 
.052 
.070 


45 
46 
47 
48 
49 


.707 
.719 
.731 
.743 

.755 


.707 
.695 
.682 
.669 
.656 


1.000 
1.036 
1.072 
1.111 
1.150 


5 
6 
7 
8 
9 


.087 
.105 
.122 
.139 
.156 


.996 
.995 
.993 
.990 
.988 


.087 
.105 
.123 
.141 

.158 


50 
51 
52 
53 
54 


.766 
.777 
.788 
.799 
.809 


.643 
.629 
.616 
.602 
.588 


1.192 
1.235 
1.280 
1.327 
1.376 


10 
11 
12 
13 
14 


.174 
.191 
.208 
.225 
.242 


.985 
.982 
.978 
.974 
.970 


.176 
.194 
.213 
.231 
.249 


55 
56 
57 
58 
59 


.819 

.829 
.839 
.848 
.857 


.574 
.559 
.545 
.530 
.515 


1.428 
1.483 
1.540 
1.600 
1.664 


15 
16 
17 
18 
19 


.259 
.276 
.292 
.309 
.326 


.966 
.961 
.956 
.951- 
.946 


.268 
.287 
.306 
.325 
.344 


60 
61 
62 
63 
64 


.866 
.875 
.883 
.891 
.899 


.500 
.485 
.469 

.454 
.438 


1.732 
1.804 
1.881 
1.963 
2.050 


20 
21 
22 
23 
24 


.342 
.358 
.375 
.391 

.407 


.940 
.934 
.927 
.921 
.914 


.364 
.384 
.404 
.424 
.445 


65 
66 
67 
68 
69 


.906 
.914 
.921 
.927 
.934 


.423 
.407 
.391 
.375 

.358 


2.145 
2.246 
2.356 
2.475 
2.605 


25 
26 
27 
28 
29 


.423 
.438 
.454 
.469 
.485 


.906 
.899 
.891 
.883 

.875 


.466 
.488 
.510 
.532 
.554 


70 
71 
72 
73 

74 


.940 
.946 
.951 
.956 
.961 


.342 
.326 
.309 
.292 
.276 


2.747 
2.904 
3.078 
3.271 
3.487 


30 
31 
32 
33 
34 


.500 
.515 
.530 
.545 
.559 


.866 
.857 
.848 
.839 
.829 


.577 
.601 
.625 
.649 
.675 


75 
76 
77 
78 
79 


.966 
.970 
.974 
.978 
.982 


.259 
.242 
.225 
.208 
.191 


3.732 
4.011 
4.331 
4.705 
5.145 . 


35 
36 
37 
38 
39 


.574 
.588 
.602 
.616 
.629 


.819 
.809 
.799 

.788 
.777 


.700 
.727 
.754 
.781 
.810 


80 
8.1 
82 
83 
84 


.985 
.988 
.990 
.993 
.995 


.174 
.156 
.139 
.122 
.105 


5.671 
6.314 
7.115 
8.144 
9.514 


40 
41 
42 
43 
44 


.643 
.656 
.669 
.682 
.695 


.766 
.755 
.743 
.731 
.719 


.839 
.869 
.900 
.933 
.966 


85 
86 
87 
88 
89 


.996 
.998 
.999 
.999 
1.000 


.087 
.070 
.052 
.035 
.017 


11.430 
14.301 
19.081 
28.636 
57.290 


45 


.707 


.707 


1.000 


90 


1.000 


.000 





NOTE. The abbreviation hyp. means "hypotenuse"; adj. means "the 
side adjacent to the angle"; opp. means "the side opposite the angle." 



362 GENEliAL MATHEMATICS 

EXERCISES 

1. A balloon B (Fig. 272) is anchored to the ground at a 
point .1 by a rope, making an angle of 57 with the ground. 
The point C on the ground directly under the balloon is 146 ft. 
from A. Assuming the rope to be straight, find 
the height of the balloon. 

Solution. Let a = height of balloon. 

Then - = tangent of 57. 

146 

But by the table, Art. 410, 

tan 57 = 1.54. 




Hence -- = 1.51. A 146/ 

Fir 272 
Solving for a. a = 224.84 ft. 

NOTK. The figure does not need to be drawn accurately, for our 
results are obtained independently of it. The solution is brief and 
depends for its accuracy upon the accuracy of the angle 57, the accu- 
racy of the length of the line A C, and the accuracy of the tangent table. 

2. The angle of elevation of an aeroplane at a point A on 
level ground is 53. The point C on the ground directly 
under the aeroplane is 315 yd. from A. Find the height of 
the aeroplane. 

3. The length of a kite string is 210yd. and the angle of 
elevation of the kite is 48. Find the height of the kite, 
supposing the line of the kite string to be straight. 

4. A pole 20 ft. in length stands vertically in a horizontal 
area, and the length of its shadow is 16.78 ft. Find the angle 
of elevation of the sun. 

HINT. Find the value of the tangent - Then look in the table 

o 

to see what angle has a tangent corresponding to the value of 

- It may be necessary for you to approximate, since the table is 

b 

not calculated for minutes. Ask your instructor to show you a more 

complete table of trigonometric ratios. 



TRIGONOMETRY 



5. A tree is broken by the wind so that its two parts form 
with the ground a right-angled triangle. The upper part makes 
an angle of 55 with the ground, and the distance on the 
ground from the trunk to the to.p of the tree is 57 ft. Find 
the length of the tree. 

6. A circular pool has a pole standing vertically at its 
center, and its top is 50 ft. above the surface. At a point in 
the edge of the pool the angle subtended by the pole is 25. 
Find the radius and the area of the pool. 

7. A ladder 35 ft. long leans against a house and reaches 
to a point 19.6 ft. from the ground. Find the angle between 
the ladder and the house and the distance the foot of the 
ladder is from the house. 

8. Measure two adjacent edges of your desk or of a rec- 
tangular table, say your study table. Find the angles that the 
diagonal makes with the edges (1) by drawing an accurate 
figure and measuring the angle with a pro- 
tractor ; (2) by use of the trigonometric ratios. 

9. The tread of a step on a certain stairway 
is 11 in. wide ; the step rises 8 in. above the next 
lower step. Find the angle at which the stairway 
rises (1) by means of a protractor and an accurate 
figure ; (2) by means of a trigonometric ratio. 

10. To find the distance across a lake (Fig. 273) 

between two points A and C, a surveyor measured off 71 ft. on 
a line EC perpendicular to A C. He ^,Z> 

then found ZC^=53. Find A C. 

11. The Washington Monument 

is 555 ft. high. How far apart are g x A C 

two observers who from points due Fjo 274 

west of the monument observe its 

angles of elevation to be 20 and 38 respectively ? (See Fig. 274.) 
HINT. Find. If. 




FIG. 273 




Then 



x + value of A C 



= tan 20 



364 



GENERAL MATHEMATICS 



*12. A man standing on the bank of a river observes that the 
angle of elevation of the top of a tree on the opposite bank is 
56 ; when he retires 55 ft. from the edge of the river the 
angle of elevation is 32. Find the height of the tree and the 
width of the river. 

*13. From the 
summit of a hill 
(Fig. 275) there are 
observed two con- 
secutive milestones on a straight horizontal road running from 
the base of the hill. The angles of depression are found to be 
13 and 8 respectively. Find the height of the hill. 




FIG. 275 



HINT. Construct TC _L CM Z . 
Let CM l = x. 



Then 



- = tan 77, (Why?) 
h 



and 



x + 1 
h 



= tan 82. (Why?) 



(1) 
(2) 



Subtracting (1) from (2), - = tan 82 - tan 77. 
Consult the table on page 36, substitute, and solve for /;. 

*14. A railroad having a hundred-foot right of way cuts 
through a farmer's field as shown in Fig. 276. If the field is 
rectangular and the measurements 
are made as shown, find the num- 
ber of square rods occupied by the 
right of way and the assessed damage 
if the land is appraised at $200 

an acre. 

FIG. 276 

.15. A ship has sailed due south- 
west a distance of 2.05 mi. How far is the ship south of 
the starting point ? How far is it west of the starting point ? 




16. From the top of a mountain 4260 ft. above sea level the 
angle of depression of a distant boat is 41. How far is the 
boat from the summit of the mountain ? 

*17. Sketch the figure and solve the right-angled triangle 
ABC when 

(e) I = 92.5, c = 100 ft. 
(f ) a = 15.2, c = 50 ft. 
(g) .1 = 40, c = 80 ft. 
(h) B = 82, c = 100 ft. 



(a) A = 30, a = 30 ft. 
. (b) B = 42, b = 60 ft. 

(c) A = 64, f, = 22 ft. 

(d) a = 35, I = 85 ft. 



411. A trigonometric formula for the area of a triangle. 

It can be shown that the area of a triangle equals half the 

product of any two sides multiplied by 

the nine of the included angle ; that is, 

ab sin A 

~2~ 

Solution. In Fig. 277 construct the 
altitude CD. 

Then T (the area) = y. (Why?) (1) 

But - = sin A (see the definition of "sine"). (2) 




Whence h=bsinA. (Why?) 

Substituting the value of 7t in (1), 

_ be sin A 
i ^ * 



(3) 



EXERCISES 

1. A boy discovers that his father's drug store completely 
covers their triangular lot and that it extends 60 ft. and 80 ft. 
on two sides from a corner. With a field protractor he measures 
the angle between the streets and finds it to be 58. He then 
tries to find the area of the lot. What result should he get ? 
* 2. Prove that the area of a parallelogram equals the product 
of two sides and the sine of the angle included between these 
two sides. 



366 GENERAL MATHEMATICS 

SUMMARY 

412. This chapter has taught the meaning of the fol- 
lowing words and phrases : scale drawing, surveyor's chain, 
steel tape, angle of elevation, angle of depression, bearing 
of a line, bearing of a point, sine of an angle, cosine of an 
angle, tangent of an angle, trigonometric ratios or trigo- 
nometric functions, solving a triangle. 

413. Scale drawings were used as a means of indirect 
measurement. 

414. A scale drawing is useful in making estimates 
of angles, lines, and areas, in getting a clear picture in 
mind of the relation of the parts that make up the figure, 
and in checking the accuracy of an algebraic solution. 
However, it is not as brief and accurate as the algebraic- 
sol ution. 

415. If two right triangles have an acute angle of one 
equal to an acute angle of the other, the ratios of their 
corresponding sides are equal. 

416. The chapter contains a table of trigonometric ratios 
of angles from 1 to 89 and .correct to three decimal places. 

417. Trigonometric ratios furnish us with a powerful 
method of solving triangles. 

418. The area of a triangle may be expressed by the 

,. , be sin A 
formula T= 

419. The area of a parallelogram equals the product of 
two sides and the sine of the angle included between these 
two sides. 



CHAPTER XV 



THEORY AND APPLICATION OF SIMULTANEOUS LINEAR 
EQUATIONS ; CLASSIFIED LISTS OF VERBAL PROBLEMS 

420. Two unknowns ; solution by the graphic method. 
In solving verbal problems it is sometimes desirable to 
use two unknowns. This chapter aims to teach three 
methods which the pupil may apply to such problems. The 
graphic method is 
shown in the dis- 
cussion of the fol- 
lowing problem: 

In a baseball game 
between the Chicago 
Cubs and the New 
York , Giants, the 
Cubs made four 
more hits than the 
Giants. How many 
hits did each team 
make? 



-1-5- 



-10 



x- 



Fir.. 278 



If we let x repre- 
sent the number of 
hits made by the Cubs 
and y the number made by the Giants, then the equation x = y + 4 
expresses the condition as set forth in the problem. 

Obviously there are any number of possible combinations such 
that the number of hits made by one team may be four more than 
the number made by the other team. This is clearly shown in the 
graph of the eqxiation x = ?/ + 4 in Fig. 278. 

367* 



3b'S 



GENERAL MATHEMATICS 



EXERCISES 

1. From the graph in Fig. 278 find the number of hits 
made by the Giants, assuming that the Cubs made 6; 8; 
10; 15; 20. 

2. Show that every point (with integral coordinates) on the 
line will give a possible combination of hits such that x y + 4. 

NOTE. By this time the student is no doubt convinced that a 
definite solution of the problem as stated is impossible, because it 
involves two unknowns and we have given but one fact. Another fact 
which should have been 
included in the problem 
is that the total number of 
hits made by both teams 
was 18. 

If we write the equa- 
tion x + y =18, expressing 
'this second fact, and study 
it by means of the graph, 
Fig. 279, we see that there 
is more than one possi- 
ble combination such that 
the total number of hits 
made is 18. 



-20 



-10 






FIG. 279 



3. Find from the graph 
in Fig. 279 the number 

of hits made by the Giants if the Cubs made 4 ; 6 ; 9 ; 12 ; 15. 

4. Show .that every point (with integral coordinates) on the 
line will give a possible solution for the equation x + y = 18. 

XOTE. We have not been able to obtain a definite solution, 
because we have been considering the two facts about the ball game 
separately. The two equations express different relations between 
the two unknowns in the ball game. This means that we must find 
one pair of numbers which will satisfy both equations, and to do 
this we must graph both equations on the same sheet to the same 
scale, as shown in Fig. 280. 



SIMULTANEOUS LINEAR EQUATIONS 369 



5. Find a point in the graph of Fig. 280 that lies on both 
lines. What does this mean ? 

6. What are the a:- values and y- values of this point ? 

7. Show that the pair of values obtained in Ex. 6 will satisfy 
both equations. 



(1) 






y 


4 





9 


5 


20 


16 



-20 




x = y + 4, (1) 
x + y = 18. (2) 



FIG. 280. THE GRAPH OF A PAIR OF SIMUL- 
TANEOUS LINEAR EQUATIONS 



The preceding exercises show that the point of inter- 
section, namely x = 11, y = 7, is on both lines, and hence 
the solution of the problem is x = 11, y = 7. Thus the 
number of hits made by the Cubs was 11, and the number 
made by the Giants was 7. 

421. Simultaneous linear equations. A pair of linear equa- 
tions which are satisfied at the same time (simultaneously) 
by the same pair of values are called simultaneous linear 
equations. The graph is a pair of intersecting straight lines. 

422. System of equations. A pair of equations like 
those in Art. 420 is often called a system of two linear 
equations. A system of linear equations having a common 



'61 it GENERAL MATHEMATICS 

solution is said to be solved when the correct values of the 
unknowns are determined. In the graphic method the 
coordinates of the point of intersection furnish the solu- 
tion. The following is a summary of the graphic method 
of solving a pair of simultaneous linear equations: 

1. Graph loth equations to the same scale. 
'2. Find the point of intersection of the two Hnrs obtained 
in 1. 

3. Estimate as accurately as possible the x-raluc find the 
y-value of this point. 

4. Check by substituting in both equation*. 

EXERCISES 

1. Solve the following systems by the graphic method and 
check each : 

M* + y=s7 ' , v2x + 3y = 23, 5y-3* = 19, 

W - ' ) } 



3* + 2^=27, 2y + 3x=13, 

^5z-4*/ = l. ( ; 5y-6x=-S. ' 

2. What difficulties did you have in finding the correct 
values for the coordinates of the points of intersection in the 
problems of Ex. 1 ? 

423. Indeterminate equations. A single equation in two 
unknowns is satisfied by an infinite (unlimited) number 
of values, but there is no one pair of values which satisfy 
it to the exclusion of all the others ; for example, the equa- 
tion x + y = 4 is satisfied by as many pairs of values as 
are represented by each distinct point on the graph of 
the equation x + y = 4. Such an equation is called an 
in determinate equation. 



SIMULTANEOUS LINEAR EQUATIONS 371 



EXERCISES 

1. Find three solutions for each of the following indeter- 
minate equations : 

(a) x + y = 7. (c) y z = 6. (e) 5 x z = 2. 
(b)m + 3n = 5. (d) 2 x - 4 ?/ = 3. (f) 3z - 4 >/ -1= 0. 

424. Contradictory equations. It sometimes happens that 
even though we have two equations in tyvo unknowns, it 
is still impossible to obtain a distinct or a unique solution, 
as is shown by the following example: 

Find two numbers such that their difference is 12 arid such 
that twice the first diminished by twice the second is equal to 14. 



-20 



-1-0 



20- 



&Z 



f 



FIG. 281. THE GRAPH OF A PAIR OF CONTRADICTORY EQUATIONS 

If we let x denote one number and y the other, then from the 
first condition, 



x - y = 12. 
From the second condition, '2 .c '2 y 14. 



(1) 

(2) 



In order to study the problem fxirther we will construct the 
graphs of (1) and (2) with reference to the same coordinate 
axes (Fig. 281). 



372 GENERAL MATHEMATICS 

EXERCISES 

1. What relation seems to exist between the two lines of 
the graph in Fig. 281 ? 

2. Are there, then, any two numbers which will satisfy the 
conditions of the problem given on the preceding page ? 

A system of equations which expresses a contradictory 
relation between the unknowns is called a system of 
contradictory, or inconsistent, equations. The graph consists 
of two (at least) parallel lines. The definition suggests 
that in a verbal problem one of the given conditions is 
not true. 

425. Identical equations. A type of problem which has no 
unique solution but admits of many solutions is illustrated 
by the following problem : 

Divide a pole 10 ft. long into two parts so that 3 times the 
first part increased by 3 times the second part is equal to 30. 

If we let x and ?/ denote the length of the two parts, the conditions 
of the problem are represented by the equations 

x + y = 10, (l) 

and 3 x + 3 y = 30. (2) 

EXERCISES 

1. Graph the equations x + y = 10 and 3 x + 3 y = 30 to 
the same scale. Interpret the graph. 

2. Divide the equation 3sc + 3?/ = 30 by 3 and compare 
the result with the equation x -\- y = 10. 

Equations like (1) and (2), above, which express the 
same relation between the unknowns, are called identical, 
dependent, or equivalent equations. Like an indeterminate 
equation, they have an infinite number of solutions but 



SIMULTANEOUS LINEAR EQUATIONS 373 

no distinct solution. Their graphs coincide. If a verbal 
problem leads to two identical equations, one condition 
has been expressed in two different ways. 

426. Outline of systems of equations and their number of 
solutions. We have seen that a linear system of equations 
in two unknowns may be 

1. Determinant and have a distinct solution. (Tlie lines 
intersect.) 

2. Contradictory and have no distinct solution. {The lines 
are parallel.} 

3. Identical and have an infinite number of solutions. ( The 
lines are coincident.} 

EXERCISES 

1. Classify the following systems according to the preceding 
outline by drawing graphs of each system : 



= , - = , 

; 6 x + 8 y = 10. ; 2 x + 3 y = 3. 

2. Could you have classified the four systems in Ex. 1 
without graphing them ? Explain. 

427. Algebraic methods of solving systems of equations in 
two unknowns. It is often difficult (sometimes impossible) 
to judge the exact values in a graphic solution. The graphic 
method helps us to see what is meant by a solution, but 
i,t is not, in general, as exact and concise a method as the 
algebraic methods which we shall now illustrate. 

428. Elimination. To solve a system without the use 
of graphs it will be necessary to reduce the two equations 
in two unknowns to one equation in one unknown. This 
process is called elimination. 



374 GENEKAL MATHEMATICS 

429. Elimination by addition or subtraction. The two 
problems which follow illustrate the method of elimination 
by addition or subtraction. 

Solve x + y = 6, (1) 

Solution. Multiplying (1) by 3 so as to make the coefficients of 
// numerically the same in both equations, 

3 x + 3 y = 18 ("j) 

2ar-3y = 2 (4) 

Adding, or =20 

Substituting 4 for .c in (1), 4 + ^ = 6. 
Solving for y, y = 2. 

4 + 2 = 6, 
Check. 8-6 = 2. 

This method is called elimination by addition. Why ? 

Solve 3;r-4y = l, (1) 

2 a- -f 5 y = 7. (2) 

Solution. Multiplying (1) by 2 and (2) by 3 so as to make the 
coefficients of x numerically equal, 

6 x - 8 y = 2 (3) 

t;., -f 15 y = -21 (4) 

Subtracting, - 23 y = 23 

Substituting 1 for // in (1), 3 x + 4 =1. 

3 a: =-3. 

a 

-c = 1. 

Hefeee y = 1, 

and x = 1. 

This method is called elimination by subtraction. Why ? 



SIMULTANEOUS LINEAR EQUATIONS 375 

430. Outline of elimination by addition or subtraction. 
To solve two simultaneous linear equations involving two 
unknowns by the method of addition or subtraction, proceed 
as follows : 

1. Multiply, if necessary, the members of the first and 
second equations by such numbers as irill make the coefficients 
of one of the unknowns numerically the same in both equations. 

2. If the coefficients have the same signs, subtract one 
equation from the other; if they have opposite signs, add 
the equations. This eliminates one unknown. 

3. Solve the equation resulting from step 2 for the unknown 
which it contains. 

4. Substitute the value of the unknown found in step 3 in 
either equation containing both unknowns and solve for the 
second unknown. 

5. Check the solution by substituting in both of the given 
equations the values found. 

EXERCISES 

Solve and check the following systems by the method of 
addition or subtraction : 

' 3 x + 2 y 7, ox 3 '// = 14, 3 x o // = 23, 

- 2z + 3y=8. " 2a; + 4//=10. '' 1 x + <j = - 35. 

4z-3?/ = -l, 7a- + 9y=-lo, 2ar + 3y = 0. 

! ' 5* + 2y=16: " 5ar-9y = -21. '' 3x + 2i/ = o. 

llx-7y=-6, . :--2y = 54 

' 



11. x = y y 



' x + y 8. 



x ?/ = , m n _ 

+ = 



in . _ 

' $(,.+11,-- 28. 3 10 



376 



GENERAL MATHEMATICS 



60 



FIG. 282 



GEOMETRIC EXERCISES FOR ALGEBRAIC SOLUTION 

13. The combined height of a tower and flagpole is 110 ft. ; 
the height of the tower is 70 ft. more than the length of the 
flagpole. Find the height of the tower and the length of 
the flagpole. 

14. A rectangular field is 25 rd. longer than it is wide. The 
perimeter of the field is 130 rd. Find the dimensions of the field. 

15. The perimeter of a football field is 320 yd. and the 
length is 10 yd. more than twice 

the width. What are its dimensions ? 

16. The circumference of a circle 
exceeds the diameter by 75 ft. Find 
the circumference and the diameter. 
(Use the formula C = - 2 y 2 d.~) 

17. Two parallel lines are cut by 
a transversal forming eight angles, 

as shown in Fig. 282. Find x and y, and all of the angles. 

18. The interior angles on the same side of a transversal 
cutting two parallel lines are (5 x 3 y) and (x + y). Their 
difference is 70. Find x and y. 

19. Two adjacent angles of a parallelogram are represented 

/o \ o 

by 3 (2m - ri) and i-g- + 4 n } , 

and their difference is 30. Find m, n, 
and the angles of the parallelogram. 

20. The difference between the 
acute angles of a right triangle is 

43. Find the acute angles. 2g3 

21. A picture frame 1 in. wide 

(Fig. 283) has an area of 44 sq. in. The picture inside the 
frame is 4 in. longer than it is wide. Find the dimensions 
of the picture. 




SIMULTANEOUS LINEAR EQUATIONS 377 

431. Elimination by the method of substitution. The 
following problem illustrates the method of elimination 
by substitution : 

Solve m -f n = 5, (1) 

2 m -5n=-ll. (2) 

Solution. Solving for i in terms of n in the first equation, 

w = 5 - n. (3) 

Substituting this value of m in (2), 

2(o-i)-5ti=-ll. (4) 

10-2-5n=-ll. 
- 7 n = - 21. 

n-=.3. . 

Substituting 3 for n in (1), m + 3 = 5, 

m = 2. 
Hence m = 22, and n = 3. 

Check. 2 + = 5, 

4 - 15 = - 11. 

The preceding method is called the method of substitution. 

432. Outline of steps in the solution of a system by the 
substitution method of elimination. To solve a system of 
linear equations containing two unknowns by the method 
of substitution, proceed as follows: 

1. Sol v<' one equation for one unknown in terms of the other. 

2. Substitute for the unknown in the other equation the 
lvalue found for it in step 1. 

3. Solve the e<juation resulting in step 2 for the unknown 
which it contains. 

4. Substitute the value of the unknown obtained in step 3 
in any equation containing both unknowns and solve for the 
second unknown. 

5. Check the solution by substituting 'in the given equations. 



378 GENERAL MATHEMATICS 

EXERCISES 

Solve the following problems by the method of substitution: 

3ff + 46 = ll, -2ar-5?/? = 17, 3m + 2n = 130, 

' oa-b = 3. ' 3x + 2><; = 2. ' 5 m - 6 n = 30. 

7s + 3y = l, 2* + 3y = 8, ^ y _ 

"llz-5y = 21. > '3z-5y = 42. 5 + 2 ' 

9* + 7-=-12, 7ar-3y=8, *' x _ // 

15*-2r=21 8 + 2y = l34. 3 2 

NUMBER-RELATION* PROBLEMS 

Solve the following problems by the method of substitution, 
and check by one of the other methods : 

9. Find two numbers whose sum is 150 and whose difference 
is 10. 

10. Find two numbers whose difference is 15 such that when 
one is added to twice the other the sum is 295. 

11. Find two numbers such that 8 times the first plus 4 
times the second equals 100, and 3 times the first plus 7 times 
the second equals 87. 

12. The quotient of two numbers is 2 and their sum is 54. 
Find the numbers. 

13. The value of a certain fraction is 4-. If 2 is added to the 
numerator and 7 to the denominator, the value of the resulting 
fraction is ^. Find the fraction. 

14. The sum of the two digits in a two-place number is 8. 
If 18 be subtracted from the number, the resulting number 
will be expressed by the original digits in reverse order. Find 
the number. 

Solution. Let u represent the digit in units' place and t the digit 
in tens' place. 

Then 10 t + u represents the original number. 

From the first condition, t + u = 8. 



SIMULTANEOUS LINEAR EQUATIONS 379 

From the second condition, 

10 1 + u - 18 = 10 u + t. (2) 

Simplifying (2), t - u = 2. (3) 

Solving (1) and (3), 2 t = 10. 

t = 5. 

Substituting 5 for / in (1), = 3. 
Therefore the number is 53. 

15. The tens' digit of a two-place number is three times 
the units' digit. If 54 be subtracted from the number, the 
difference is a number expressed by the digits in the reverse 
order. Find the number. 

16. If a two-digit number be decreased by 13, and this dif- 
ference divided by the sum of the digits, the quotient is 5. 
If the number be divided by one fourth of the units' digit, 
the quotient is 49. Find the number. 

433. Summary of methods of elimination. This chapter 
has taught the following three methods of solving a system 
of simultaneous linear equations : 

1. The graphic method. 

2. Elimination by addition or subtraction. 

3. Elimination by substitution. 

EXERCISES 

Some of your classmates may be interested to learn a fourth 
method called elimination by comparison. Turn to a standard 
algebra and report to your classmates on this method. 

MIXTURE PROBLEMS 

Solve the following problems by any method : 
1. A grocer has two kinds of coffee, one worth 300 per 
pound and another worth 200 per pound. How many pounds 
of the 30-cent coffee must be mixed with 12 Ib. of the 20-cent 
coffee to make a mixture worth 240 per pound? 



380 GENERAL -MATHEMATICS 

2. A grocer makes a mixture of 20-cent nuts and 32-cent 
nuts to sell at 28 a pound. What quantities of each grade of 
nuts must he take to make 60 Ib. of the mixture ? 

3. How much milk testing 5% butter fat and cream testing 
25% butter fat must be mixed to make 30 gal. that test 22% 
butter fat ? 

4. How much milk testing 3.7% butter fat and cream testing 
25.5% butter fat must be mixed to make 15 gal. that test 20% 
butter fat ? 

5. "What numl)er of ounces of gold 75% pure and 85% pure 
must be mixed to give 10 oz. of gold 80% pure ? 

G. An alloy of copper and silver weighing 50 oz. contains 
5 oz. of copper. How much silver must be added so that 10 oz. 
of the new alloy may contain ^ oz. of copper ? 

7. The standard daily diet for an adult requires about 75 g. 
of protein and 100 g. of fat. Mutton (leg) contains 19.8% 
protein and 12.4% fat. Bread (average) contains 9.2% protein 
and 1.3% fat. Find how many grains each of bread and mutton 
are required to furnish the required amount of protein and fat 
in a standard ration for one day. 

HINT. Let x and y represent the number of grains required of 
mutton and bread respectively. 

Then 0.198 x + 0.092 y = 75, (1) 

and 0.121 x + 0.013 y = 100. (2) 

Solve the equations (1) and (2) simultaneously. 

If x or y turns out negative, we know that it is not possible to make 
up a standard diet out of the two foods mentioned. 

8. The table on the following page gives the amounts 
of protein and fats in the various foods often used in the 
daily diet. 



SIMULTANEOUS LINEAR EQUATIONS 381 



FOOD 


PER CENT 
OF PROTEIN 


PER CENT 
OF FAT 


Mutton (le (r ) 


19.8 


12.4 


Beef (roast) 


22.3 


28.6 


Pork (chops) 
Esfsrs . 


16.6 
13.4 


30.1 
10.5 


Bread (average) 


9.2 


1.3 


Beans (dried) 
Cabbage 


22.5 
1.6 


1.8 
0.3 


Rice 


8.0 


0.3 









Find three pairs of food combinations that will make a 
standard diet and determine the number of grams required of 
each in the following list : 

(a) Mutton and rice. (f ) Bread and cabbage. 

(b) Eggs and rice. (g) Beans and cabbage. 

(c) Bread and eggs. (h) Bread and beans. 

(d) Pork and bread. ( i ) Beef and bread. 

(e) Pork and beans. (j) Beef and rice. 

434. Systems of equations containing fractions. The fol- 
lowing list of problems offers no new difficulties. The 
student merely needs to remember to remove the fractions 
in each equation by multiplying through by the L.C.M. of 
the denominators in each, thus reducing each equation to 
the standard form ax + by = c, where a represents the co- 
efficient of x, b the coefficient of y, and c the constant term. 

5 
: 6' 



3 

x y 



or 



4 3 12 

The first equation may be written 

x + y=i: 
Similarly, (2) reduces to x y = \^ 



(1) 
(2) 



(3) 
Why? 



382 GENERAL MATHEMATICS 

EXERCISES 
Reduce to the standard form, and solve : 



1. 

2. 
3. 


X + If 


a; 3 y 




m + 1 1 


3 

2x + 


3 

'/ 3 ?/ 4- x 


> 

5 


4. + l 2' 
m-1 1 


3 
7 5 


4 


4 

1 


71-1 4 

-f 3 .s + 4 5 


5 
jr. y 


3 

, ^ + y 


!' 

2 


5. 5 10 2 ' 
f> * ~ 


2 


3 

./ i ./ 


3 


Zt .) .s 


7 3 

0.3 x + 0.7 y = 5.7, 
2 x 4 y = 4. 


3 

x 4- 5 


1 !~ 



435. Linear systems of the type -+- = c\ work 

. y 

problems. In the problems of Art. 434 we have seen the 
advisability of reducing each of the equations in a system 
to the standard form by eliminating the fractions and 
collecting similar terms. There are some problems, how- 
ever, in which it is advisable to solve without eliminating 
the fractions. An example will illustrate what is meant. 

Two pipes can fill ^ of a cistern if the first runs 2 hr. 
and the second runs 3 hr., but if the first runs 3 hr. and the 
second runs 2 hr., they can fill ^ of the cistern. 

Solution. Let x = the number of hours it will take the first 

pipe alone to fill the cistern, 
. 
and y = the number of; hours it will take the 

second pipe alone to fill the cistern. 

Then - = the part of the cistern the first pipe can 

fill in 1 hr. 

and - = the part of the cistern the second pipe 

y can fill in llir. 



SIMULTANEOUS LINEAR EQUATIONS 383 
From the first condition. 

; + r^r (1) 

From the second condition, 

- + - = 7! (2) 

x y lo 

Multiplying (1) by 3 and (2) by 2, 
6 9 _ 27 

and - + - = ~ (4) 

x y 15 

5 27 28 
Subtracting, - = - (5) 

(Note that this is a linear equation in one unknown.) 
Solving, y - G, 

x = 5. 
Problems like the preceding are called work problems. 

WORK PROBLEMS 

1. A and B can build a fence in 4 da. If A works 6 da. and 
quits, B can finish the work in 3 da. In how many days can 
each do the work alone ? 

2. A tank can be filled by two pipes one of which runs 3 hr. 
and the other 7 hr., or by the same two pipes if one runs 5 hr. 
and the other 6 hr. How long will it take each pipe alone to 
fill the tank ? 

3. A mechanic and an apprentice receive $4.40 for a job of 
work. The mechanic works 5 hr. and the apprentice 8 hr. 
Working at another time, and at the same rate per hour, the 
mechanic works 10 hr. and the apprentice 11 hr., and they 
receive &7.30. What are the wages per hour for each? 



384 GENERAL MATHEMATICS 

4. Solve the following problems without getting rid of the 
fractions, and check : 

1 1 _5 47_15 
H 7; > . i 

* x y 6 3 x y 4 

'J-i-i 'M = i. 

x i/ (y x y 

2 3 = 19 11 _ 9 = 26 

2 //i n 15 , s f 5 

' A . = i ' 13_3 = 8 

m n 5 s t 5 

436. Review list of verbal problems. The following 
problems review types studied in earlier chapters. In actual 
practice many problems may be solved by using either one 
or two unknowns. In general it is advisable to use one un- 
known, but sometimes it is easier to translate the problem 
into algebraic language if two unknowns are used. It will 
be helpful if some member of the class will show the two 
methods in contrast. 

MOTION PROBLEMS 

1. A crew can row 8 mi. downstream in 40 min. and 12 mi. 
upstream in 1 hr. and 30 min. Find the rate in miles per hour 
of the current and the rate of the crew in still water. 

Solution. Let x = the rate of the crew in still water, 

and y = the rate of the current. 

Then, if we express the rates in miles per hour, 



x-y = = S. (2) 

(TO 

Adding, 2 x = 20. 

Hence x = 10, the rate of the crew in still water, 

and y = 2, the rate of the current. 



SIMULTANEOUS LINEAR EQUATIONS 880 

2. A boatman rowed 10 mi. upstream and 4 mi. back in 
3' lir. If the velocity of the current was 2 mi. per hour, what 
was his rate of rowing ? 

3. The report of a gun traveled 367yd. per sec. with tin- 
wind and 353 yd. per sec. against the wind. Find the velocity 
of sound and the rate at which the wind was blowing. 

4. An ae'roplane flying with a wind blowing at the rate of 
2 mi. an hour consumes 2 hr. 30 min. in going a certain dis- 
tance and 2 hr. 42 min. 30 sec. in returning. Find the distance 
and the rate of the aeroplane in still air. 

BEAM PROBLEMS 

NOTK. Beam, or lever, problems which involve two unknowns 
are readily solved by means of the laws of leverages and forces 
discussed in Art. 233. 



5. Two boys carry 
a bag of coal weigh- 
ing 25 Ib. by hanging M 




it on a pole 8 ft. long k I n ,__ J~ 

at a point 2 ft. from I 

the middle of the pole. +x JJ5 + y 

How much of the -load F 284 

does each boy carry ? 

Solution. Let x and y represent the number of pounds carried 
by each boy. 

Then, using M, Fig. 284 (the middle of the pole), as the turning 
point, 

4 y - 50 - 4 x = 0. (Why ?) 

From which // x = 12. .1. (1) 

x + y = 25. (Why ?) (2) 

Adding (1) and (2), 2 ?/ = 37.f>. 

r~i8f. 

Substituting for y in (2), 18;} ./ = (i\. 



386 GENERAL MATHEMATICS 

6. Two weights balance when one is 14 in. and the other 
10 in. from the fulcrum. If the first weight is increased by 
2 Ib. and placed 10 in. from the fulcrum, the balance is main- 
tained. Find the number of pounds in each weight. 
" --3Two weights balance when one is 12 in. and the other 
is 10 inT^em the fulcrum. The balance is maintained if the 
first weight isriKived 3 in. nearer the fulcrum and if 3 Ib. is 
subtracted from the s&soml. Find the weights. 

8. An iron bar 6 ft. loug^^^ighing 20 Ib. is used by two 
boys, one at each end, to carry a Io9ji-oi^oO Ib. How many 
pounds must each boy carry if the load hangs~2~ttr-feaio_the 
right end ? (Consider the weight of the entire bar as hanging 
at the middle of the bar.) 

9. A wagon bed 12 ft. long is loaded with 20 cakes of ice 
weighing 250 Ib. The bed extends 2 ft. over the front axle of 
the running gears and 3 ft. behind the rear axle. Find the load 
supported by each axle. 

10. The material in a 30-foot bridge weighs 3600 Ib. The 
bridge supports two loads : 700 Ib. at 3 ft. from one end, and 
1500 Ib. at 5 ft. from the other end. Find the loads borne by 
the two supports. 

11. Three men have to carry an oak beam 15 ft. long weigh- 
ing 250 Ib. Two of the men lift at the ends of an iron bar 
placed crosswise beneath the beam, and the third man lifts at 
the rear end of the beam. Where must the iron bar be placed 
in order that each man will carry one third of the load ? 

* RECREATION PROBLEMS 

12. A man upon being asked the age of his two sons, 
answered, " If to the sum of their ages 18 be added, the result 
will be double the age of the elder ; but if 6 be taken from the 
difference of their ages, the remainder will be equal to the age 
of the younger." Find the age of each. 



SIMULTANEOUS LINEAR EQUATIONS 387 

13. In a guessing game the leader says, " If you will add 
10 years to your age, divide the sum by your age, add 6 to the 
quotient, and tell me the result, I will tell you your age." How 
did he find it ? 

14. A baseball team has played 40 games, of which it has 
won 28. How many games must it win in succession in order 
to bring its average of games won up to 0.750 ? 

15. A girl has worked 12 problems. If she should work 13 
more problems and get 8 of them right, her average would be 
72%. How many problems has she worked correctly thus far ? 

16. Two bicycle riders ride together around a circular track, 
one along the outside edge, where the radius of the circle is 
R, and the other along the inside edge, where the radius is r. 
One revolution of the pedals carries the former's bicycle 20 ft. 
and the latter's 25 ft. Write a formula expressing the differ- 
ence between the number of pedal revolutions made by the two 
cyclists in going around the track once ; five times ; n times. 

17. If 10 marbles of one size are dropped into a bucket of 
water, and the water rises a, inches, and 15 equal marbles of 
another size are dropped into the same bucket, and the water 
rises b inches, write a formula showing how many times larger 
one of the first size is than one of the second size. 

18. An automobile .tire when fully inflated has a radius of 
18 in. Owing to a leakage of air, this is reduced to 17 in. Indi- 
cate how many more revolutions per mile are necessary because 
of the leakage. If the original radius is R and the reduced 
radius r, what is the formula which could be used to calculate 
the difference of revolutions per mile ? 

19. Divide $183 into two parts, so that ^ of the first part 
shall be equal to y 3 ^ of the second part. 

20. Each of two brothers wanted to buy a lot valued at 
$240. The elder brother said to the younger, " You lend me 
| of your money, and I can purchase the lt-" "But," said the 



088 GENERAL MATHEMATICS 

younger brother, ''you lend me | of your money, and I ran 
purchase the lot." How much money did each have '.' 

21. A group of boys bought a touring car. After paying for 
it they discovered that if there had been one boy more, They 
would have paid $30 apiece less, but if there had been our 
boy less, they would have paid $60 apiece more. How many 
boys were there, and what did they pay for the car? 

22. The Champion American League team one year (1914) 
won 46 more games than it lost. The team standing second 
played 153 games, winning 8 less than the first and losing 
9 more than the first. Find the number of games won and 
lost by each team. 

23. It is said that the following problem was assigned by 
Euclid to his pupils about three centuries B.C. : " A mule and 
a donkey were going to market laden with wheat. The mule 
said to the donkey, ' If you were to give me one measure, 
I would carry twice as much as you ; if I were to give you one 
measure, our burdens would: be equal.' What was the burden 
of each ? " 

* MISCELLANEOUS PROBLEMS 

24. A bar 30 in. long is balanced by a 40-pound weight at one 
end and a 32-pound weight at the other end. Find the position 
nl' the support. 

25. A man has a f 2000 exemption frOm income tax, but pays 
at the rate of 6% on the rest of his income. He finds that after 
paying income tax his actual income is $3410. On what amount 
does he pay the 6% tax ? 

26. A chemist has the same acid in two strengths. If 16 1. 
of the second are mixed with 241. of the first, "the mixture is 
42% pure, and if 61. of the first are mixed with 41. of the 
second, the mixture is 43% pure. Find the per cent of purity 
of each acid. Problems like this are often given as practical 
problems. Find out from some chemist why it is not practical. 



SIMULTANEOUS LINEAR EQUATIONS 389 

27. After a strike a corporation decided to raise the wages 
of each laborer from .? to y by the formula y = tux + i>, where 
ra and b are to be determined by the facts that a man who 
made $2 is to receive $2.30,. and one who made $2.20 is to 
receive $2.41. Find ra and b, also the new wage of a man 
who formerly received $3, $4, $4.20. 

28. At what market price must one-year 5% bonds be offered 
for sale in order that the buyer, by holding them until maturity, 
nitiy make 6% on his investment? 

HINT. The profit made must come from two sources : the interest 
on the par value, which is $5, and the excess of the maturity value 
over the price paid for the bonds. If x is the price paid, then 
100 - x + 5 = 0.06 x. 

SUMMARY - 

437. This chapter has taught the meaning of the fol- 
lowing words and phrases : simultaneous equations, linear 
systems of equations, indeterminate equations, contradic- 
tory equations, identical equations, elimination. 

438. This chapter has taught the following methods of 
solving a system of equations in two unknowns: 

1. Solution by graph. 

2. Solution by addition or subtraction. 

3. Solution by substitution. 

439. The student has been taught how to solve systems 

involving fractions, and systems of the type - -f - = c. 

x y 

440. The following types of verbal problems have been 
introduced: geometric problems, number-relation problems, 
mixture problems, work problems, motion problems, beam 
problems, and recreation problems. 



CHAPTER XVI 

GEOMETRIC AND ALGEBRAIC INTERPRETATION OF 
ROOTS AND POWERS 

441. Introductory work; square root. The following 
exercises are introductory to the work of the chapter. 

EXERCISES 

1. What number multiplied by itself equals 9? 16? 121? 
169? x 2 ? if? 

2. How many answers are there to each part of Ex. 1? 
(Why?) 

3. One of the two equal factors of a number is called the 
square root of the number. What is the square root of 49 ? 

4 4y 2 

of 64? of 625? ofar 2 ? of 4^? of J? o^-J 

9 9y* 

4. The positive square root of a number is indicated by a 
sign ( V ) called the radical sign, and either the radical sign 
alone (V ) 01> the radical sign preceded by the plus sign 
(+ V ) means the positive square root of the number under- 
neath the sign. The number underneath the radical sign is 
called the radicand. The negative square root is indicated by 
the radical sign preceded by the minus sign ( V ) With 
the preceding definitions in mind give the value of the fol- 
lowing: V25; Vl6; VlOO; -Vl21; V(X25 ; -Vl44; Vj; 



9 ' \25iy 2 

5. Express the following statement by means of a formula: 
A number y equals the square of another number x. 

390 



INTERPRETATION OF ROOTS AND POWERS 391 

6. If x = 1 in the formula y = ar 2 , what is the value of y? 
If x = 2, what is the value of y ? 

7. Calculate the corresponding values of y in the formula 
y = x' for each of the following values ofic: 1 ; 2 ; -(-3; 

Q. i 1 . i 2 . 2 

~~ t> y ~7~ 7)~" j T "3* 5 "*""" Q 

8. Fill in the proper values in the following table of 
squares and square roots for use in the next article. 



X 


i 


- 1 




2 




- 2 


3 




4 


5 6 




b 









I 


i 


i 


y 


i 


i 




i 




4 


















. 


i 












































X 


4 





T 




t 


s 


i i 


: j 


- i 


1 


V 






t 


\ 


i 


- 





y 














i. 




























































































































































-I ^ 






































































































































^ 






































> 


































y 




r 


































~J 




i 


































J 






\ 
































/ 












































\ 






























/ 






































j 








\ L 






























/ 








L 






























1 
















































V 


























^ 












\ 
























^ 














A 
























/ 














m 






















J 
















^ 






















/ 


















1 ^ 


















^ k 




































^i 
































































\^ 










. ' 






























"s^ L 








* 






























^ ^ 








^^ 






























tSi 








^ ^ 


















- JF 














"%i. - 







5 J 
















'T 


" ~ 5 " 


n ~ 










-' 






r-i- p 


i; 




-, 












5 


-:t~ 












t 








_t : 














i^-^ 


f" 



FIG. 285. DEVICE FOR FINDING SQUARES AND SQUARE ROOTS 

442. Graph of y = x*\ a device for finding squares and 
square roots. The values of the preceding table have 
been plotted in Fig. 285. Values for x were laid off 



392 GENERAL MATHEMATICS 

horizontally on the 2>axis, and the corresponding values 
for y vertically on the ^-axis. The points were then con- 
nected by a smooth curve, as shown. This curve serves 
as, a device for determining squares and square roots, as 
we shall now see. 



EXERCISES 



1. Determine by the graph in Fig. 285 the square root of 1 ' : 
9; 2,0; 22; 3; 2. How many answers' do you obtain for the 
square root of 9 ? 4 ? 25 '.' for the square root of each number 
shown? 

2. By means of the graph in Fig. 285 find the square of 2; 
1.4; 2.2; 2.4; 3.3; 5.6; 3.9; 1.7. 

3. Check your results for Ex. 2 by squaring the numbers 
given. The squares should be approximately those you found 
by means of the graph. 

4. How would you make a graph that would give you 
squares and square roots more accurately than the graph in 
Fig. 285 ? 

443. A positive number has two square roots. The graph 
shows that the square root of 4 equals either -j- *2 or 2 ; 
that is, there are two answers for the square root of a 
positive number. Thus, (3) (3) =9, as also does 
( -H 3) ('+ 3 ). Note the symmetry of the curve in Fig. 285 
and see if you can explain it. 

444. Quadratic surd. The indicated square root of a 
number which is not a perfect square is called a quadratic 

* i\l : for example, v3, v20, V.r. 



445. Quadratic trinomial. Trinomials like a 2 -f 2 ab + b 2 
and .?-' 2 2 .ry + y 2 are of the second degree, and are called 
ijuadratic trinomials. The word " quadratic " comes from 



INTERPRETATION OF ROOTS AND POWERS 393 



the old word " quadrature," which means a geometric 
square ; hence quadratic means that a 2 4- 2 ab -f 5 2 and 
./ 2 '2xy + y 2 are of the second degree. They are the 



a 
ab 


b b 2 


2 


ab 




ab 



ab 




a + b 

FIG. 286. GEOMETRIC REPRESENTATION OF (a + b) 2 . 

squares of a + b and ^ y respectively, as we have already 
seen earlier in this book. In Fig. 286 the geometric square 
of a + b is shown. 

EXERCISES 

1. See if you can point out where we have worked with 
problems dealing with the square of binomials like a + b 
and x y. 

2. Where, earlier in the book, have we learned how to take 
the square root of trinomial squares by factoring ? 

3. Find the value of the following : 



Vra 2 + 2 mn + n* ; V4 "a? + 12 xy + 9 if ; V9 a 2 - 12 6 +T7X 

*446. Square roots of algebraic polynomials and arith- 
metical numbers. We have already seen in Chapter IX 
how the square root of a trinomial is found by reversing the 
process of squaring a binomial : that is, .by finding one of 
the two equal factors of the trinomial. The same process 
may be explained geometrically if the exercises given on 
page 394 are carefully solved. 



394 



GENERAL MATHEMATICS 



ab 



ab 



FIG. 287 



ILLUSTRATIVE EXERCISES 

1. Find the square root of 16 x~ + 40a-y + 25 y 2 

Solution. If this trinomial is a perfect square of some binomial, it 
may be illustrated by Fig. 287, in which the side of the largest square 
obtained by inspection and corresponding to a 2 is 4 x. Therefore 
the side of each rectangle corresponding to , 

each ab is 4 x, and the area correspond- 
ing to 2 ab + b 2 must be 40 xy + 25 y' 2 . The 
problem therefore consists in determining 
the width of the strip which we are adding 
on two sides and which corresponds to the 
b of the formula. In this case b is 5 y. Now 
5 y may be obtained by dividing 40 xy by 
the sum of 4 x and 4 x, or 8 x. Hence 
doubling the term already found (4 x) the 
result 8 x serves as a divisor for determin- 
ing the next term. Fig. 287 shows that we double 4 x because 
8# is approximately the combined length of the strip to which 
we are adding. This is illustrated more clearly in the next problem. 

2. Two boys were asked to stake out a square plot of 
ground with an area .of 4225 sq. ft. 5 
What is the length of a side ? 

Solution. The boys' thinking about the 
problem might take some such form as 
follows : 

(a) It is obvious that we can make it 

at least 60 by 60. We shall suppose that 60 

this is constructed. See the square with 

unbroken lines (Fig. 288). This uses up p IGi 288 

:5600 sq. ft., leaving 625 sq. ft. We can add 

to the square already constructed by adding to two sides and still 

keep it a square. 

(b) The combined length of the edges to which we are adding 
is 120 ft. Hence the approximate length of the strip added is 
120 ft. Why approximate? 

(c) 120 is contained five times in 625 (with a remainder). 



INTERPRETATION OF ROOTS AND POWERS 395 



(d) If we make the strip 5 ft. wide, the total length will be 125 
(for one strip will be 60 ft. and the other 65 ft.). 

(e) 125 is contained exactly five times in 625. 

(f) Hence the square must be constructed so as to be 65' by 65'; 
that is, the square root of 4225 is 65. 

3. Find the square root of the polynomial a 2 + 2 ab + b 2 
+ 2ac + 2bc + c 2 (see Fig. 289). 

Solution. The side of the largest square is a^ therefore the trial 
divisor is 2 a. The width of the first strip is b, therefore the divisor 
is 2 a + b. Multiplying by b and subtracting the 
remainder gives 2 OK + 2 be + c 2 . The length of 
the square now constructed is a + b. The edge 
to which we are adding is 2 a + 2 b units long 
(trial divisor). 2 a + 2 b is contained c times 
in 2 ac + 2 be. If we make the strip c units 
wide, the total length of the strip to which we 
add is 2 a + 2 b + c (complete divisor). (Why ?) 
Multiplying and subtracting, the remainder is 
zero. The side of the total square is a '+ b + c, or 




FIG. 289 



Va 2 + 2 ab + b z + 2 ac + 2 be + c 2 = a + b + c. 
The work may be arranged as follows : 
Largest square, a 2 a 2 + 2 ah + b 2 + 2 ac + 2 be 4- c 2 [ a + b + c 



2 ab + b* 

2 ah + b 2 



First trial divisor, 2 a 
First complete divisor, 2 <i + l> 
Second trial divisor, 2 a + 2 b 
Second complete divisor, 2 a + 2 I + c 



2 ac + 2 be + c 2 
2 ac + 2 be + c 2 



4. Find the first digit in the square root of 177,2^1. 

Solution. To determine this first digit we must remember (a) that 
the square of a number of one digit consists of one or two digits, the 
square of a number of two digits consists of three or four digits, the 
square of a number of three digits consists of five or six digits, and so 
on ; (b) that the number of digits in the integral part of the square 
of a number is twice as large or one less than twice as large as the 
number of digits in the integral part of the given number. This 
suggests the following device for determining the number of digits 



396 GENEKAL MATHEMATICS 

in the integral part of the square root of a number. Beginning at 
the decimal jioint, mark off toward the left groups of two digits 
each. Then the number of digits in the square root will be the 
same as the number of groups. Thus, since 177,241 is made up of 
three groups of two digits (17'72'41'), the square root of 177,241 
contains three digits in its integral part. We are thus able to esti- 
mate the largest square as 400 (that is, the first digit is 4) and then 
proceed as in Ex. 3. The work may be arranged as follows : 

17 72'41 1 400 + 20 + 1 

16 00 00 

First trial divisor, 800 I 1 72 41 
First complete divisor, 820 1 1 64 00 
Second trial divisor, 840 
Second complete divisor, 841 
Therefore Vl77241 = 421. 

*447. Steps involved in finding square roots. The follow- 
ing steps were used in Exs. 1~4, above ; the student should 
study them carefully. 

1. Estimate the largest square in the number. 

2. Double the root already found for a trial divisor. 

3. Divide the first term of the remainder by the trial 
ilirt'xor, placing the quotient as the next term of the root-. 

4. Annex the term just found to the trial divisor to form a 
complete divisor and continue the process until the other fcn/i* 
f>f the root are found. 

EXERCISES 

Find the square roots of the following polynomials : 

1. <>' 2 + 2ab + b 2 . -"3. 4 4 +-4 8 + 9a 2 + 4 0+4. 

2. 16z* + 2xy + 9//'. 6. x 4 - 2x 3 + 3x 2 - 2 x + 1. 

3. 49 if - 14 yz + z\ 7. 1 - 4 a + 6 . 2 - 4 8 + \ 
4. x*+2x 8 + 3x 2 + 2r + l. 8. 

9. x 6 + 4 ax* - 2 aV + 4 aV - 4 <>\r + a 6 . 
10. 9 + 12 // + 6 f + 4 ?/ + 4 if + if. 



INTERPRETATION OF ROOTS AND POWERS 397 



11. ** + 8. 

, '.9, 

12. 7 + 6a 
4 

13. 576. 

14. 9025. 

15. 51,529. 



_ 16 4 16. 61,504. 

~ x 42' 17. 57,121. 
6 a* 2 _8a 3 18. 2. 

2 # 2 . NOTE. Write 2.00WOO and 

proceed as in Ex. 4, Art. 446. 

19. 3. 

20. 3.1416. 



448. Table of roots and powers. In practical situations 
it is convenient to use a table of roots and powers. There 
are a number of very useful tables in textbook or leaflet 
form, and the student is now in a position where he c,an 
easily learn how to use them. A very simple table of 
roots and powers is submitted on page 398. It will fre-- 
quently prove a great convenience to the student in his 
work on the following pages. 

449. The theorem of Pythagoras. If we study the follow- 
ing exercise carefully we shall discover a well-known 
geometric theorem which will be useful in later work. 

EXERCISE 

Construct a right triangle, making the 'sides including the 
right angle 3 and 4 units long respectively (see AAJ'>( ', Fig. 290). 
Using the same unit, find the length of J />'. 
On each side draw a square and divide 
each square into unit squares. Counting 
these squares, find how the square on the 
hypotenuse compares with the sum of the 
squares on the other two sides. 

The preceding exercise illustrates 

,. , Fro. 290 

the familiar theorem of Pythagoras: 

/// it right triangle the sum of the squares on the sides inclml- 
l n < i the- -right angle is-^ual to the .squaw on. tlie ' 




398 



GENERAL MATHEMATICS 



TABLE OF ROOTS AND POWKKS 



No. 


Squares 


Cubes 


Square 
Roots 


Cube 
Roots 


No. 


Squares 


Cubes 


Square 
Roots 


Cube 
Roots 


1 


1 


1 


1.000 


1.000 


51 


2,601 


132,651 


7.141 


3.708 


2 


4 


8 


1.414 


1.259 


52 


2,704 


140,608 


7.211 


3.732 


3 


9 


27 


1.732 


1.442 


53 


2,809 


148,877 


7.280 


3.756 


4 


16 


64 


2.000 


1.587 


54 


2,916. 


157,464 


7.348 


3.779 


5 


25 


125 


2.236 


1.709 


55 


3,025 


166,375 


7.416 


3.802 


6 


36 


216 


2.449 


1.817 


56 


3,136 


175,616 


7.483 


3.825 


7 


49 


343 


2.645 


1.912 


57 


3,249 


185,193 


7.549 


3.848 


8 


64 


512 


2.828 


2.000 


58 


3,364 


195,112 


7.615 


3.870 


9 


81 


729 


3.000 


2.080 


59 


3,481 


205,379 


7.681 


3.892 


10 


100 


1,000 


3.162 


2.154 


60 


3,600 


216,000 


7.745 


3.914 


11 


121 


1,331 


3.316 


2.223 


61 


3,721 


226,981 


7.810 


3.936 


12 


144 


1,728 


3.464 


2.289 


62 


3,844 


238,328 


7.874 


3.957 


13 


169 


2,197 


3.605 


2.351 


63 


3,969 


250,047 


7.937 


3.979 


14 


1% 


2,744 


3.741 


2.410 


64 


4,096 


262,144 


8.000 


4.000 


15 


225 


3,375 


3.872 


2.466 


65 


4,225 


274,625 


8.062 


4.020 


16 


256 


4,096 


4.000 


2.519 


66 


4,356 


287,496 


8.124 


4.041 


17 


289 


4,913 


4.123 


2.571 


67 


4,489 


300,763 


8.185 


4.061 


18 


324 


5,832 


4.242 


2.620 


68 


4,624 


314,432 


8.246 


4.081 


19 


361 


6,859 


4.358 


2.668 


69 


4,761 


328,509 


8.306 


4.101 


20 


400 


8,000 


4.472 


2.714 


70 


4,900 


343,000 


8.366 


4.121 


21 


441 


9,261 


4.582 


2.758 


71 


5,041 


357,911 


8.426 


4.140 


22 


484 


10,648 


4.690 


2.802 


72 


5,184 


373,248 


8.485 


4.160 


23 


529 


12,167 


4.795 


2.843 


73 


5,329 


389,017 


8.544 


4.179 


24 


576 


13,824 


4.898 


2.884 


74 


5,476 


405,224 


8.602 


4.198 


25 


625 


15,625 


5.000 


2.924 


75 


5,625 


421,875 


8.660 


4.217 


26 


676 


17,576 


5.099 


2.962 


76 


5,776 


438,976 


8.717 


4.235 


27 


729 


19,683 


5.196 


3.000 


77 


5,929 


456,533 


8.774 


4.254 


28 


784 


21,952 


5.291 


3.036 


78 


6,084 


474,552 


8.831 


4.272 


29 


841 


24,389 


5.385 


3.072 


79 


6,241 


493,039 


8.888 


4.290 


30 


900 


27,000 


5.477 


3.107 


80 


6,400 


512,000 


8.944 


4.308 


31 


961 


29,791 


5.567 


3.141 


81 


6,561 


531,441 


9.000 


4.326 


32 


1,024 


32,768 


5.656 


3.174 


82 


6,724 


551,368 


9.055 


4.344 


33 


1,089 


35,937 


5.744 


3.207 


83 


6,889 


571,787 


9.110 


4.362 


34 


1,156 


39,304 


5.830 


3.239 


84 


7,056 


592,704 


9.165 


4.379 


35 


1,225 


42,875 


5.916 


3.271 


85 


7,225 


614,125 


9.219 


4.396 


36 


1,296 


46,656 


6.000 


3.301 


86 


7,396 


636,056 


9.273 


4.414 


37 


1,369 


50,653 


6.082 


3.332 


87 


7,569 


658,503 


9.327 


4.431 


38 


1,444 


54,872 


6.164 


3.361 


88 


7,744 


681,472 


9.380 


4.447 


39 


1,521 


59,319 


6.244 


3.391 


89 


7,921 


704,969 


9.433 


4.464 


40 


1,600 


64,000 


6.324 


3.419 


90 


8,100 


729,000 


9.486 


4.481 


41 


1,681 


68,921 


6.403 


3.448 


91 


8,281 


753,571 


9.539 


4.497 


42 


1,764 


74,088 


6.480 


3.476 


92 


8,464 


778,688 


9.591 


4.514 


43 


1,849 


79,507 


6.557 


3.503 


93 


8,649 


804,357 


9.643 


4.530 


44 


1,936 


85,184 


6.633 


3.530 


94 


8,836 


830,584 


9.695 


4.546 


45 


2,025 


91.125 


6.708 


3.556 


95 


9,025 


857,375 


9.746 


4.562 


46 


2,116 


97.336 


6.782 


3.583 


96 


9,216 


884,736 


9.797 


4.578 


47 


2,209 


103,823 


6.855 


3.608 


97 


9,409 


912,673 


9.848 


4.594 


48 


2,304 


110,592 


6.928 


3.634 


98 


9,604' 


941,192 


9.899 


4.610 


49 


2,401 


117,649 


7.000 


3.659 


99 


9,801 


970,299 


9.949 


4.626 


50 


2,500 


125,000 


7.071 


3.684 


100 


10,000 


1,000,000 


10.000 


4.641 



INTERPRETATION OF ROOTS AND POWERS 399 

This theorem is one of the most famous theorems of 
geometry. Centuries before Christ the Egyptians used a 
rope divided by knots so that its three lengths were in 
the ratio 3:4:5. This rope was used in land surveying 
and also in the orientation of their temples. In fact, 
we read of professional " rope fasteners " (surveyors ?). 
Furthermore, the proof of the theorem itself has always 
appealed to the interest of mathematicians. When we shall 
have advanced in our study of mathematics it will be 
possible for the student to find many proofs of this theorem 
that he can understand. The earliest general proof is 
credited to Pythagoras, who lived about 500 B.C. 

The student has probably found this theorem to be the 
basis for one of the most useful rules of arithmetic. The 
proof given in arithmetic classes is usually that given in 
the exercise above. However, a general proof demands 
that we prove the theorem independent of the accuracy 
of the figure (that is, independent of the measurements 
and constructions involved). We shall presently give 
such a proof. The exercises which follow are intended 
to review the material necessary to establish this proof. 

EXERCISES 

1. In Fig. 291 A A EC is a right triangle, right-angled at C, with 
CD AB. Re view the proof which shows that AADC^ A ABC. 

c b 

2. Prove that in Fig. 291 - = - 

and that b 2 = cm. 

b; 

3. Review the proof which shows 

that ABDC-^ A ABC. A\ 

4. Prove that in Fig. 291 - = - 

O n 41 




5. Show by using Exs. 2 and 4 that a 2 + tf = <?. 



400 



( i ION EKAL MATHEMATK s 



450. Theorem of Pythagoras proved. No doubt the 
student now sees that the theorem of Pythagoras is 
proved by Exs. 1~5, aboye. We 
shall, however, set up the prooi' 
for reference. l>, 



(liven the right triangle A /'><'. 
right-angled at C, to prove that the 
square on the hypotenuse is equal 
to the sum of the squares on the 
sides including the right angle. In terms of Fig. 292 this 
means to prove that c 2 = a 2 -\- l>~. 




Proof 



STATEMENTS 



REASONS 



In Fig. 292 draw CI)AJi 
and letter the figure as shown. 



rr, 

Then 



and 



(1) 

(2) 



In (1) and (2) I- = me and 
= nc. (3) 



By adding the two equations 
in (3), 

a- + b- = me + ne. (4) 
a 2 + ft 2 =e(jn + n). (5) 

But m + c. ((5) 



.-. n- + //- = ,--. 



(7) 



Because if in a right triangle 
a line is drawn from the vertex 
of a right angle perpendicular to 
the hypotenuse, either side about 
the right angle is a mean pro- 
portional between the whole hy- 
potenuse and the segment of the 
hypotenuse adjacent to it. 

Because when four quantities 
are in proportion the product of 
the me'ans equals the product of 
the extremes. 



Addition axiom. 
By factoring out c. 
The whole is equal to the sum 
of all its parts. 
Bv substitution. 




PYTHAGORAS 



402 GENERAL MATHEMATICS 

HISTORICAL NOTE. Pythagoras (c. 569 u.c.-c. 500 B.C.), the second 
of the great philosophers of Greece, is said to hate " changed the study 
of geometry into the form of a liberal education." After some wander- 
ings, he founded the famous Pythagorean School at Croton, a Dorian 
colony in the south of Italy. Here enthusiastic audiences composed 
of citizens of all ranks, especially the upper classes, crowded to hear 
him. It is said that the women went to hear him in direct violation 
of a law against their public appearance. 

' Pythagoras divided his audiences into two classes : the Proba- 
tioners (or listeners) and the Pythagoreans (or mathematicians). 
After three years in the first class a listener could be initiated 
into the second class, to whom were confided the main discoveries 
of the school. 

The Pythagoreans formed a brotherhood in which each member 
was bound by oath not to reveal the teachings or secrets of the 
school. Their food was simple, their discipline severe, and their 
mode of life arranged to encourage self-command, temperance, purity, 
and obedience. 

The triple interwoven triangle, or pentagram (star-shaped regular 
pentagon), was used as a sign of recognition, and was to them a symbol 
of health. It is related that a Pythagorean while traveling fell ill and, 
although carefully nursed by a kind-hearted innkeeper, was unable 
to survive. Before dying, however, he inscribed the pentagram star 
on a board and begged his host to hang it up outside. This the host 
did ; and after a considerable length of time another Pythagorean, 
passing by, noticed the sign and, after hearing the innkeeper's story, 
rewarded him handsomely. One motto of the brotherhood was : 
"A figure and a step forwards; not a figure to gain three oboli." 

The views of society advocated by the brotherhood were opposite 
to those of the democratic party of Pythagoras's time, and hence 
most of the brotherhood were aristocrats. For a short time the 
Pythagoreans succeeded in dominating affairs, but a popular revolt 
in 501 B.C led to the murder of many prominent members of the 
school, and Pythagoras himself was killed shortly afterwards. 

Though the brotherhood no longer existed as a political party, the 
Pythagoreans continued to exist a long time as a philosophical and 
mathematical society, but to the end remained a secret organization, 
publishing nothing, and thus leaving us little information as'to the 
details of their history. See Ball's "A History of Mathematics," p. 19. 



EXERCISES 

1. The base and altitude of a right triangle (Fig. 293) are 
6 and 8 respectively. What is the length of the hypotenuse ? 
(Use the theorem of Pythagoras.) 

2. How long must a rope be run from the 
top of a 16-foot tent pole to a point 20 ft. from 
the foot of the pole ? 

3. A baseball diamond is a square a side of 

which is 90 ft. What is the length of a throw 

_b IG. t'f> 

from " home " to " second " ? 

4. Find the formula for the diagonal of a square whose 
side is s. Use this formula to determine the diagonal when 
s = 10 ; when s = 15. 

5. Prove from the Pythagorean theorem that a = Vc 2 6 2 
and translate the equation into words. 

6. Prove also that b = Vc 2 a 2 and translate the equation 
into words. 

7. A ladder 20ft. long just reaches a window 15ft. above 
the ground. How far is the foot of the ladder from the foot 
of the wall if the ground is level ? 

8. The hypotenuse of a right triangle is 35 ft. and the 
altitude is 21 ft. Find the base. 

9. Using the formula of Ex. 6, find the value of a when 
c = 22 and b = 20. 

10. A tree standing on level ground was broken 24 ft. from 
the ground, and the top struck the ground 18 ft. from the stump, 
the broken end remaining on the stump. How tall was the tree 
before breaking ? 

11. Construct on squared paper a right triangle, using 
the following pairs of numbers for the base and altitude 



404 



GENERAL MATHEMATICS 



respectively : 1 and 1 ; 1 and V2 ; 2 and 2 ; 2 and 3 ; 4 and 4 ; 
4 and 5 ; 1 and 5 ; 2 and 5 ; 3 and 5 ; 12 and 1. 

HINT. Use the line segment you obtained for the first part. 

12. Calculate for each part of Ex. 11 the length of the 
hypotenuse. 

451. The theorem of Pythagoras furnishes a method of 
constructing with ruler and compasses the square root of a 
number. Exercises 11 and 12, Art. 450, suggest a method 
of finding the square root of a number by 
means of ruler and compasses. The method 
is illustrated by the following exercise : 

Construct the square root of 42. 

The following study (analysis) of the exer- 
cise will help us to understand the problem. 

Suppose that we have the figure constructed ; 
that is, let us imagine that Fig. 294 is the required 
figure and that AB is the required length V42. 

Now a and b can be of various lengths provided 
a- + tf = 42. (Why?) Let us suppose that CB - 6 ; 
then how long is l>'l We know that o(> + tr 42 
would be the equation from which the value of b can be found. 
It is clear that b would have to equal A 7/ 6. 

Then the problem merely becomes one of learning how to con- 
struct V6. Some members of the class may already know how 
to do this, but we shall proceed with our analysis. 

"Imagine another triangle, A'B'C' (Fig. 295), so con- 
structed that the hypotenuse turns out to be V6 and so 
that B'C' is 2 units long; then A 'C' must equal V2. Why? 

Our problem finally reduces, then, to a problem of 
constructing V2. If we can find this geometrically we 
can solve the original exercise, as our analysis has shown. 

We already know how to construct v2 by constructing a right 
triangle with the two legs about the right angle equal to 1. Then 
the hypotenuse equals V2. Why? 





INTERPRETATION OF ROOTS AND POWERS 405 

We then reverse our analysis as follows : 

(a) Construct V2 as indicated above. 

(b) Construct a second right triangle with a base of v'2 units long 
and an altitude 2 units long. Its hypotenuse will equal Vti. Why? 

(c) Construct a third right triangle whose base is \/6 units long 
and whose altitude is 6 units long. Its hypotenuse will be \/42 units. 

Why? 

EXERCISE 

Construct with compasses a line segment equivalent to each of 
the following: V6 ; VTT ; A/27 ; Vl43; V214 ; 3 VJJ ; 2 V2. 

452. Mean proportional construction a method for find- 
ing square roots. We shall now see that our mean pro- 
portional construction (Art. 374) furnishes us with an easy 
method of constructing square roots. 

EXERCISES 

1. Review the construction (Art. 374) for finding a mean 
proportional between two line segments a and b (Fig. 296). 

2. Construct the mean proportional a 

between 4 and 9 ; 4 and 16. ft 



3. Review the proof for the statement 

.. , T , FJG. 296 

that a mean proportional between two line 

segments a and b equals the square root of the product of a and //. 

The preceding exercises suggest that the mean propor- 
tional construction furnishes a method for finding the square 
root of a number. For example, 

Find the square root of 12. 

On squared paper find the mean propor- . 
tional of two factors of 12, for example, 2 / 
and 6. The mean proportional x (Fig. 297) {_ 
is the square root of 12, for K 2- 

1 -. Why? FIG. 297. MEAN PROPOR- 

y 6 TIONAL METHOD OF FIND- 

x 2 = 12^_ Why? ING THE SQUARE ROOT OF 

Whence x = Vl2. Whv? A NUMBER 



406 GENERAL MATHEMATICS 

EXERCISE 

Construct the square root of 21 ; 6 ; 5 ; 18 ; 42 ; 84 ; 66 ; 
76. Compare the results with the table of Art. 449. Your 
results ought to approximate the second decimal place. 

453. Large numbers under the radical signs. When the 
number under the radical sign is large, the various geo- 
metric constructions for finding square roots are neither 
convenient nor, in general, sufficiently accurate. In this 
case it is of advantage to reduce the given quadratic surd 
to an equivalent expression which has a smaller number 
under the radical sign. Suppose we wish to find the value 
of V5056. The square root is at once evident if we resolve 
the number into two equal groups of factors ; thus : 



V5056 = V(2 2 3 . 7) (2 2 3 . 7) = V84 84 = 84. 

Even when the number is not a perfect square, factor- 
ing will often enable the student to- find its square root 
much more easily, as will be shown later. 

EXERCISES 

Find the following indicated square roots : 

1. V576. 3. V484. 5. V3600. 

2. V1296. 4. V1089. 6. Vl936. 

454. The square root of a product. The preceding exer- 
cises show that the square root of the product of several 
factors, each of which is a square, may be found by taking 
the square root of each factor separately, as in the follow- 
ing examples: 

1. V9- 25=V9V25 = 3- 5=15. 




INTERPRETATION OF ROOTS AND POWERS 407 

This is true because 9 25 can be written as the product 
of two groups of equal factors (3 5) (3 5). Hence, by the 
definition of a square root, (3 5) is the square root of 9 25. 

2. Vl6 xY = Vl6 V^ Vp = 4 a*f. 

This is true because 16 afy 6 may be written as the product 
of two groups of equal factors (4 a^?/ 3 ) (4 z 2 ?/ 3 ). Hence 
4 a; 2 ?/ 3 is the square root of IGa^y 6 . 

The preceding exercises show that the square root of a 
product is obtained by finding the square root of each factor 
separately and then taking the product of these roots. That 

is, in general, _ _ 

v a b = v a Vft. 

This principle may be used to simplify radical surds in 
the following manner. Suppose we wish to find the value 

of V11858. Then _ 

V11858 = V2 11- 11 7-7 



= 77 V2. 

By the table of roots, V% = 1.414. 

Then 77(1.414) = 108.878. 

Hence V11858 = 108.878. 

It will be helpful to observe the following: 

(1) The principle enables us to simplify the radicand to 
a point where we can easily find the root by the table or 
by several geometric constructions. 

(2) A quadratic surd is in its simplest form when the 
number under the radical sign does not contain a perfect 
square factor. 

In general, if the expression under the radical sign 
contains a factor which is a square, this factor may be 
removed by writing its square root before the radical sign. 



-108 GENERAL MATHEMATICS 

EXERCISES 

Change the following so as to leave no factor which is a 
square under the radical sign : 



1. Vl2. 


_6. V2S. 


11. 


2. V40. 


7. V75. 


12. 


3. VlS. 


8. V125. 


v 13. 


4. V50. 


^9. VlOS. 


- 14. 


' 5. V72. 


10. Vo*. 


* 15. 




455. Value of memorizing square roots of certain numbers. 
Exercises 1~9, Art. 454, suggest the manner in which the 
square roots of a few small numbers, like 2, 3, 5, are 
made to do service in finding the roots of many large 
numbers. In fact, many students in other fields find that 
memorizing these numbers, which occur again and again 
in their problems, increases their efficiency. 

EXERCISES 

From the table of roots we know that V2=1.414, A/3=1.732, 
and V 5 = 2.236. Using these facts, compute each of the fol- 
lowing correct to two decimal places : 

1. V75. 4. V72. 7. V50 + V75-V6. 

2. V80. 5. V98. 8. 2 V32 + V72 - Vl8. 

3. V48. 6. V363. 9. V45-\/|. 

D 

456. The square root of a fraction. A fraction is squared 
by squaring its numerator and its denominator separately 

9 

and indicating the product thus : x - = . Hence, to 

bob 

extract the square root of a fraction, we find the square 
root of its numerator and denominator separately. 
For example, VT\ = J, since % % = f V- 



INTERPRETATION OF ROOTS AND POWERS 409 

EXERCISES 

Find the square roots of the following accurate to two places 
(use the table for square roots): 

1. f. 2. f. 3. TV 4. f. 5. f. 6. V-. 7. _*_. 8 . _2 y . 

457. Rationalizing the denominator of a fraction. It 
probably has been noticed that the calculation of those 
problems of the preceding lists in which the denominator 
is a square is far easier than when this is not the case. 
This may be illustrated as follows: 



[2 _ V2 



1.414 
1.732 



Dividing 1.414 by 1.732 is not an easy division. In fact, 
most people would find it impossible to do mentally. On 

the other hand, J? = _ = ?dii? = Q.816 involves an easy 

M9 V9 8 
mental division. 

It is of interest to note that the denominator of a 
fraction can always be made a square merely by multi- 
plying numerator and denominator by the proper factor ; 



, Jf = J? = ^ = 
i o \ 9 o 



thus, = = = = 0.816. Note that the 

o 

division - is an easy mental division compared with 
o 

the division ' , which we had above. 



A\ r e may summarize this method of taking the square 
root of a fraction as follows : 

1. To find the square root of a fraction, change the fraction 
into an equivalent fraction whose denominator is a perfect 
square. 



410 GENERAL MATHEMATICS 

2. The square root of the netv fraction equals the square 
root of its numerator divided by the square root of its 
denominator. 

3. If desired, express the result in simplest decimal form. 

The process of changing a radical expression so as to 
leave no denominator under a radical sign is called ration- 

alizing the denominator. Thus, -yj - = - V3 ; \J- = -^a- 

1 3 o \ a a 

EXERCISES 

1. Find the value of the following square roots (find values 
approximately accurate to the second decimal place) : 

(a) J. (b) f . (c) 1 (d) 1 (e) f . (f) f (g) T V 

2. Rationalize denominators of the following, and simplify : 



3. What is the value of the expressions in Ex. 2 from (e) 
to (j), inclusive, if a = 3 and b = 2 ? 

458. Addition and subtraction of surds. Sometimes the 
arithmetic in a problem may be simplified if the surds are 
combined into one term. Thus, 

V20 + V45 = 2 V5 + 3 V5 

= 5V5, or5(2.236) 
= 11.180. 

By adding 2 \/5 + 3 V5 just as we add 2_- 4 plus 3 4, we 
need only to look up the \/5, whereas V20 + V45 calls for 
two square roots. 



INTERPRETATION OF ROOTS AND POWERS 411 

In the following list simplify each expression as far as 
possible without using approximate roots ; that is, leave 
your result in indicated form. Practically this is often 
better than finding an approximation, for in this manner 
you submit results that are absolutely accurate. It leaves 
the approximation to the next person, who may find as 
many decimal places as the needs of his particular problem 
demand. 

EXERCISES 

1. Vl08 + V75 + Vl2. 8. 

O 

2. 2V98-VT8. 9. Vf + 

3. 6 V288 - 4 Vl8 + Vl28. 10. Vf- 

4. 5 V432 - 4 V3 + Vl47. n. 1+^ 

3 

5. 2 V27 + 3 V48 -*3 V75. 12. - f + 

6. 3 V20 + 2 Vl25 - Vl80. 13. ^/T _ 2 V2 + 

4- V- 14. 4 V28 + 3 Vo5 - Vll2. 



459. Multiplication and division of quadratic surds. These 
two processes will be treated briefly. In elementary mathe- 
matics there are few verbal problems which involve this 
process. Further, the principles involved do not offer any- 
thing new for us. 

Thus, to divide V2 by V5, we may write this in the 

form of a fraction and proceed as we ordinarily do 

when finding the value of a fraction which involves 
quadratic surds. 

The rule in multiplication is equally familiar. The 
equation Vab=\ / aVb may be read just as well from 
right to left, and we have VaV5=VaJ. Thus, V2V5 is 
precisely the same as VlO. 



n-2 GENERAL MATHEMATICS 

EXERCISES 

1. Find the product of the following: 

(a) V3 V27. (d) V(5Wl2a; 8 ~. 

(b) vW* 8 . (e) V|VH-_ 

(c) V3V5. (f) V|Vf Vf Vl 

( g ) ( V2 + Va - Vs)( V2 - Vs + Vs). 

Solution. We multiply each term of one polynomial by each term 
of the other and simplify the result as follows : 

V2 + V - V5 
VQ - Va + Vs 
2 + Ve - Vio 
- Ve - a + Vis 
+ VTo + Vis - 5 



2 Vlo - 6 

(h) (2 Vr + Vs - Vs)( Vs). 

(i) (4V2-2V3 + V5)(2 V3). 

(j) (Vs - 3 V2 + VTo)(Vs). 

(k) (3 V2 + 4 Vs)(2 V2 - S Vs). 

(i) ( Vs - VI + Vs)(V3 + VI - Vs). 

2. Divide, and express the result in simplest form : 

(a) 1 by VS. (c) V6 by Vs. 

(b) 24 by Vs. (d) 2 Vl2 - 5 VlS by Vs. 

460. Fractional exponents another -means of indicating 
roots and powers. Thus far we have not used a fraction 
as an exponent, and evidently it could not be so used 
without extending the meaning of the word "exponent." 

Thus, a? means x x x, but X s evidently cannot mean 
that x is to be used as a factor one half of a time. 



INTERPRETATION OF ROOTS AND POWERS 413 

It is very important that all exponents should be gov- 
erned by the same laws; therefore, instead of giving a 
formal definition of fractional exponents, we shall lay down 
the one condition that the laws for integral exponents shall 
be generally true, and we shall permit the fractional ex- 
ponent to assume a meaning necessary in order that the 
exponent laws shall hold. 

Since we agree that ^ x^ x, we see that a?* is one of 
the two equal factors of x ; that is, x^ is the square root 
of #, or we may write y? = V#. 

Similarly, since X s X s x 5 = z, X s is the cube root of x ; 

\ 3/ 

that is, X s = v x. 

Again, x* x* x' 5 = X s = xt. This means that X s is the 

2 3/ 

cube root of 2 ; that is, x* = v x 2 . 

This discussion is sufficient to show that the fractional 
exponent under the laws which govern integral exponents 
takes on a meaning which makes a fractional exponent 
just another way of indicating roots and powers ; that is, 
the denominator indicates the root, and the numerator indi- 
cates the power. 

Thus, 8^ means to take the cube root of 8 and square 
the result, or square 8 and take the cube root of the result. 
In either case the final, result is 4. Again, 10* (or 10- 6666 ) 
means to take the cube root of 100, which, by the table of 
Art. 449, is 4.641. Note that a common-fraction exponent 
may appear as a decimal fraction. 

From this it will be obvious that the student needs only 
to become familiar with the new method of writing roots 
and powers. The following brief list of problems presents 
precisely the same ideas as those in the preceding list 
dealing with surds ; only the form is different. 



414 GENERAL MATHEMATICS 

EXERCISES 
1. Write in simplest form : 

(a) 4* + 9* + 16* + 25* + 36*. 

(b) 1* + 8* + 64* + 0*. 

(c) 64* + 9* + 16* + (- 32)* - (- 27)1 

(d) 24* + 54' - 6*. 

(e) 18* + 32* - Vl28 + V2. 

(f) (^*+8*-(f) 4 +(50)*. 

(g) (81)* - 2 (24)* + V28 + 2 (63)*. 



2. Multiply: 

(a) z*z*. (c) 10 2 - 5 10-. 125 10 1 - 25 . 

(b) 10* 10* - 10*. (d) 10 3 - 6250 10- 3750 . 10- 0625 . 

3. Translate (c) and (d), Ex. 2, into expressions with common 
fractions as exponents and estimate how large the numbers 
would be if we had a way of finding the result. 

NOTE. 10 - 8 " 5 means 10^, or the eighth root of 1000. Another 
way to look at it is that 10 is to be raised to the 375th power and 
the thousandth root taken. These ideas are of very great importance 
in getting a clear understanding of the chapters on logarithms and 
the slide rule. In these chapters we shall learn how to find a root 
(say the 15th) just as easily as the square root. 

461. Zero exponents. Under the laws which govern 
integral exponents a = 1, as is shown by the following : 
5 -=- a 5 = a. (By the Division Law.) 
o 5 -5-o 6 = l. (The quotient of any number divided 

by itself is 1.) 
Hence a = l. (By the equality axiom.) 

Thus the zero power of any number (except zero) is 1. 
Thus, 15 =1; (560)= 1; (-6a:) =l; 10 = 1. 



462. Negative exponents. If x~ s obeys the same law as 
integral exponents, then 

_ Q Q Q -| 

-3- X = , or ; 

1 X 6 X 6 X 6 

that is, x~ s 

Similarly, if we multiply the numerator and denominator 

x~ a 1 

of - - by x a , we obtain x~ a = 

Tin-, ~~> looAo-^jL, v-'-mf 

J^ = 10,000. 

See if you can state in simple language the meaning of 
a negative exponent. 

EXERCISES 

Simplify : 

1. 10- 2 x 10 3 - 125 x 10 - 0625 x 10- 1 - 03675 . 

2. 10^ x lo" x (56) x 10*. 

3. (39) (169)"*. 6. 16- 2- 4 . 9. x m x n . 

4. 1000 (100)"* 7. 1000- IO- 3 . 

i in y b . y c 

5. 2 X x 2-* x 2. 8. 144~* 24. 

11. X m H-X n . 

12. (+625)*; (-125)-*. 

13. (x 2 ) 3 ; (x 3 ) 4 ; (x ) 2 . 
HINT, (z 2 ) 3 = x 2 x z x 2 . 

14. (IO 2 ) 2 ; (IO 2 ) 8 ; (IO 2 ) 4 . 

15. (10- 125 ) 2 ; (10 - 0625 ) 4 ; (10- 125 ) 8 . 

16. (x m ) n ; translate the formula (x m } n = x mn into an alge- 
braic rule. 

17. -v/x^; ^x 8 ; ^x 5 . 

NOTE. The 3 in vx means "find the cube root." 



416 



GENERAL MATHEMATICS 



18. 

19. Since (x m ) n ^ x mn , what is 

m 

Translate the formula Vic = x" into an algebraic rule. 

20. Find the value of : 

(a) VlO^. (b) A/10"-' 75 , (c) -v/H) 3 -" 250 . (d) -x/lO 5 - 8750 . 

463. Cube of a binomial. We shall now see how certain 
other short cuts in finding powers may be illustrated and 
explained. 

INTRODUCTORY EXERCISES 

1. Find the cube of x -f- y by first finding the square of 

x -f y and multiplying the result by x + y. 

2. By multiplication find (a + &) 3 . 

3. Find the cube of (x y); of (a &). Compare these 
cubes with the results of Exs. 1 

and 2. 

4. It will be helpful to your class- 
mates if you will make a set of blocks, 
as shown in Fig. 298, in order to 
show that 

(x -f- 7/) 3 = x 3 + 3 x 2 y + 3 xif + ?/ 3 - 
How many blocks are needed ? 

5. Find the following cubes, doing 
as many as you can mentally : 

(a) (c + dy. (c) (c-df. (e) (a-4-27/) 8 . (g) (2x + 

(b) (m + nf. (d) (m-nY- (f) (2 x - y) s . (h) (2x-3y) s . 

Exercises 15 show that the cube of a binomial is equal 
to the cube of the first term, plus three times the square of 
the first term multiplied by the second, plus three times the 
first term multiplied by the square of the second, plus the 
cube of the second term. 





SIR ISAAC NEWTON 



418 GENERAL MATHEMATICS 

HISTORICAL NOTE. The first use of positive and negative frac- 
tions as exponents is found in a book written by the great English 
mathematician and physicist, Sir Isaac Newton (1642-1727). He 
discovered the binomial theorem and numerous laws of physics ; 
for example, the law of gravitation, the law about lenses and prisms, 
and the explanation of the rainbow. Among his numerous books are 
" Universal Arithmetic " (really an algebra) and " Principia " (one of 
the greatest books of all times). 

The biography of Xewton (see Ball's " A Short History of Mathe- 
matics," pp. 328-362, and Cajori's "A History of Elementary Mathe- 
matics," pp. 238-240) is very interesting and inspiring. As a boy 
he was expected to be learning how to take care of his father's 
farm, but he spent much of his time studying and trying mechan- 
ical experiments. Thus, we read of his constructing a- clock run 
by water which kept very fair time. His mother noticing this 
sensibly resolved to send him to Cambridge. Here followed a brilliant 
career of thirty-five years as student and teacher. As a professor it 
was his practice to lecture publicly once a week, and then only from 
half an hour to an hour at a time. In the week following he gave 
four hours of consultation to students who wished to discuss the 
results of the previous lecture. It is said that he never repeated a 
course and that one course began at the point where the preceding 
course had ended. The result of his second study during this period 
has dazed master minds who have attempted to understand all that 
Newton accomplished. Perhaps you will later agree with some of 
the following tributes to him : 

Nature and Nature's laws lay bid in night : 

God said, " Let Newton be ! " and all was light. POPE 

There, Priest of Nature ! dost thou shine, 

Newton ! a king among the kings divine. SOCTHEY 

Tafcing mathematics from the beginning of the world to the time when 
Newton lived, what he had done was much the better half. LEIBNITZ 

I don't know what I may seem to the world, but as to myself, I 
seem to have been only as a boy playing on the sea-shore, and diverting 
myself in now and then finding a smoother pebble or a prettier shell 
than ordinary, whilst the great ocean of truth lay all undiscovered 
before me. NEWTON 



INTERPRETATION OF ROOTS AND POWERS 419 



464. Cube roots. The equation y = $ asserts that y 
is the cube of .r, or, inversely, x is the cube root of y. 
If the equa- 
tion is graphed 
we shall ob- 
tain a curve 
analogous to 
the curve for 
squares and 
square roots 
which may be 
used to find 
cube roots and 
cubes. We pro- 
ceed to find 
corresponding 
values for y 
and x in order 
that we may 
plot sufficient 
points for the 



-x- 



-y- 



curve. 



FIG. 299. GRAPH OF y - X s 



EXERCISES 

1. What is the value of y in the equation y = y? when x 
equals ? when x = + 1 ? when x 1 ? when x = + 2 ? 
when x = 2 ? 

2. Calculate values as in Ex. 1 and fill the blank spaces of 
the following table. If the curve is not obvious, expand the 
table until you have enough points to draw the curve. 



X 

y 





+ 1 
+ 1 


+ 2 
+ 8 


-2 
-8 


3 


4 


6 


6 


7 


i 


i 





f 


1 


&. 


1. 





27 















420 GENERAL MATHEMATICS 

From this table we may obtain the curve in Fig. 299. 
One small square vertically represents 1 unit, and 5 small 
squares horizontally represent 1 unit. 

From this curve we can read off, approximately, the 
cube or the cube root of any number; thus the cube of 
2.2 is seen to be about 10.5 (by the table actually 10.64); 
the cube root of 13 is seen to be a little over 2.4 (see 
the table for values accurate to two decimal places). More 
accurate results can be obtained by drawing the curve to 
a large scale. 

465. Cube roots of arithmetical numbers. An arithmetical 
method for finding cube roots based on an algebraic for- 
mula for (a + 6) 3 could be devised. But this method is 
seldom used, because cube roots as well as higher roots 
are more quickly found by means of logarithms. This 
method will be taught in the next chapter. In the mean- 
time the student may for all practical purposes use the 
table in Art. 449. 

Furthermore, we could devise analogous rules and curves 
for fourth roots, fifth roots, fourth powers, and so on, but 
these too are more readily found by logarithms. 

466. Indicating higher roots. By means of an index figure 
the radical sign is made to indicate other roots than square 
roots. 

Thus the cube root of 8 or one of its three equal factors 
is written V8 = 2. The fourth root of 16 is written Vl6 = 2. 
The 3 in v 8 is the index of the root. 

Any expression which contains an indicated root is called 
a radical expression. 

The principles of reducing surds to simpler forms, dis- 
cussed in detail for quadratic surds, may be applied to 
higher indicated roots. 



EXERCISES 

1. Simplify the following (remove any factor which is a 
perfect power of the degree indicated by the index): 

^32; -\/64; ^64 x 6 /; "\/48 4 ; A/16 a 

2. Add and subtract as indicated: 

(a) A/16 + -^54 - -^250 + -v 

(b) -v/54 + -v/128 + -v/1024 + 




SUMMARY 

467. This chapter lias taught the meaning of the fol- 
lowing words and phrases : square root of a number, 
quadratic surd, radical sign, radicand, quadratic trinomial 
square, index. 

468. The graph of the equation y = x 2 was used as a 
device for finding squares and square roots. 

469. A positive number has two square roots ; thus, 
V4 = + 2 or - 2. 

470. The square of the sum or difference of two numbers 
may be found by the formula 

(a A) 2 = a 2 2 db + ft 2 . 
This formula was illustrated geometrically. 

471. To find the square root of a perfect square tri- 
nomial: Extract the square roots of the two perfect square 
terms and connect them by the sign of the remaining term. 

472. The square of a trinomial consists of the sum of the 
squares of its terms plus twice the product of each term by 
each succeeding term. By remembering this rule the square 
roots of some polynomials may be found by inspection. 



422 GENERAL MATHEMATICS 

473. The chapter taught a method of finding the square 
root of algebraic polynomials and arithmetical numbers. 

474. The chapter includes a simple table of square 
roots and cube roots. 

475. Quadratic surds may often be simplified by apply- 
ing the principle vW>=VaVi. 

476. We find the square root of a fraction by dividing 
the square root of the numerator by the square root of 

the denominator; that is, A 7 = - 

: b V 6 

477. 'Rationalizing the denominator simplifies the cal- 

rn /~ 

culation ; that is, -yj is more difficult than -=- 

478. When the same number occurs as the radical in 
a series of terms, the terms may be combined by the 
rule for adding similar terms. This usually simplifies 
the calculation. 

479. The theorem of Pythagoras was proved. 

480. The theorem of Pythagoras furnishes a method of 
constructing the square root of a number. 

481. The mean proportional construction furnishes 
another method of finding the square root. 

482. A fractional exponent is another method of indicat- 
ing roots and powers; thus, x% means -$2?. The numerator 
indicates the power, and the denominator the root. 

483. a is defined as 1. 

484. A number with a negative exponent is defined so 
as to be equal to the reciprocal of the same number with 

a positive exponent; that is, a~ 5 = 



INTERPRETATION OF ROOTS AND POWERS 423 

485. The cube of a binomial may be found by the 
following formula: 

(a + 6) 3 = a 3 + 3 a 2 6 + 3 a 



486. Cube roots may be found by the table, graph, or 
more easily by logarithms and the slide rule (the last two 
methods will be shown in the next two chapters). 



CHAPTER XVII 

* LOGARITHMS APPLIED TO MULTIPLICATION, DIVISION, 
ROOTS AND POWERS, AND VERBAL PROBLEMS INVOLV- 
ING EXPONENTIAL EQUATIONS 

LOGARITHMS 

487. Labor-saving devices. In Chapter IV we showed 
how extensive calculations even with only four or five 
place numbers are apt to become laborious and, in some 
cases, inaccurate and involving unnecessary steps. We 
showed how to minimize the inaccuracy and how some of 
the unnecessary steps may be eliminated, especially with 
regard to the processes of multiplication and division by 
the so-called " abbreviated method." But in many cases 
the work remains long and tedious, even with the use of 
these abbreviated methods. 

In Art. 449 will be found a table of powers and roots 
which are given for the purpose of saving time and labor. 
Scientific books include similar tables which help to save time 
and conserve our energy. Other labor-saving devices com- 
monly used are adding machines, cash registers, graphs, etc. 

One of the greatest labor-saving devices by which diffi- 
cult problems may be readily solved is the method of 
calculation by logarithms. This chapter will be devoted 
to a simple explanation of this method. If the student 
will study the chapter carefully and solve the problems 
correctly, he will get a foundation in logarithmic work 
that will be very helpful in subsequent work. 

424 



LOGARITHMS 425 

488. Two methods of multiplying. We are already 
familiar with the two methods of multiplying illustrated 
by the following examples: 

100 x 1000 = 100,000. 

10 2 x 10 8 = 10 5 

= 100,000. 

The student will observe that the product is the same 
in both examples, but that the method used in the second 
has reduced the operation of actually multiplying the two 
numbers to a simple problem in the addition of exponents. 

Although the numbers multiplied here are in each case 
powers of 10, the method will hold for other bases as well. 
However, we shall consider only the base 10 in our sub- 
sequent discussion since it is the one commonly used. 

Find the product of the following pairs of numbers by 
the two methods given above : 

10 and 100. 10,000 and 100,000. 

1000 and 1000. 1000 and 1,000,000. 

489. Powers of 10. From the preceding exercises it is 
clear that we can multiply together numbers which are 
integral powers of 10 merely by adding the exponents of 
these powers. Since every positive number may be expressed 
exactly or approximately as a power of 10, we may obtain 
the product of any two numbers in a similar manner by 
adding the exponents of the powers of 10 which equal the 
respective numbers. For example, we may multiply 17.782 
and 3.162 by adding the exponents of the powers of 10, 
which equal 17.782 and 3.162 respectively. 

This raises two important questions: (1) What powers 
of 10 equal 17.782 and 3.162? (2) What is the value 
of 10 when raised to the sum of these two powers ? It is 



426 GENERAL MATHEMATICS 

possible for us to work out a table which will give us the 
powers of 10 which equal 17.782 and 3.162. The table 
below shows how the different values are obtained. The 
table is not complete, as will be shown later, but it contains 
the approximate values of several integral and fractional 
powers of 10 which we need at this point. 

We know that 10 = 1, 10 1 = 10, 10 2 = 100, 10 3 = 1000, and so on. 
\Vr can find the value of 10- 5 as follows : 

10- 5 = 10* = VlO = 3.162 (approx.). 

From these values the other values in the table are easily found, 
as the student can verify. 

TABLE OF POWERS OF 10 
10' = 1.0000 



I = V8.162 = 1.7782 
10- 5 = 10z = VlO = 3.1623 
100.75 _ (ioi.5)2 = v'31.62 = 5.6234 
10 1 = 10.0000 

lQi-25 = lOi . 100.25 = i7.7 8 o 

lQi-5 _ i i . 100.5 _ 31.623 

lOi- =10* 10<>-'5 = 56.234 

10* = 100.000 

102.25 _ i i . iQi.25 _ 177.82 

10 2 - 5 = lOi . 101.5 _ 316.23 

jO-2.75 _ 1 1 . 101-75 _ 562.34 

10 =1000.00 

We may now resume the solution of the problem proposed 
above, namely, multiplying 17.782 by 3.162. We can see 
by referring to the table that 17.782 = 10 1 - 25 and that 
3.162 =10- 5 . Hence, 17.782 x 3.162 = 10 1 - 25 xlO- 5 = 10 1 - 78 , 
which, by referring to the table, we see is equal to 56.234 
(this product is accurate to the second decimal place). 



LOGARITHMS 427 

EXERCISE 

Check the preceding result by actually multiplying 17.782 
by 3.162, and account for the difference in results. Is there a 
significant difference ? Is the result obtained by actual multi- 
plication accurate to more than two decimal places ? 

490. Logarithms ; notation for logarithms. In the equa- 
tion 17.782 =10 1 - 25 the exponent 1.25 (which indicates 
the power to which 10 must be raised to give 17.782) is 
called the logarithm of 17.782 to the base 10. 

Thus the logarithm of a number to the base 10 is tlie 
exponent of the power to which 10 must be raised to equal 
that number. From now on we shall assume that the base 
is 10 when we speak of the logarithm of a number. The 
symbol for logarithm is log. Thus, log 1000 = 3 is read " the 
logarithm of 1000 equals 3," the base 10 being understood. 

EXERCISE 

By means of the table of powers of 10 in Art. 489 find log 1 ; 
log 10 ; log 100 ; log 1.78; log 316.23. 

491. A logarithm is an exponent. The student will do 
well to remember the two ways of thinking of an expo- 
nent ; for example, in the equation 10 2 = 100 the 2 can be 
thought of (a) as the exponent of the power to which 10 
must be raised to equal 100 ; that is, 10 2 = 100 ; (b) as the 
logarithm of 100 to the base 10; that is, 2 = Iog 10 100. 

EXERCISES 

Read the following in two ways : 

1. 10 1 = 10. 2. 10 2 = 100. 3. 10 8 = 1000. 4. 10 4 = 10,000. 



428 GENERAL MATHEMATICS 

492. Characteristic ; mantissa. A glance at the table of 
Art. 489 will show that each exponent of 10 (each loga- 
rithm of the corresponding numbers to the right) may con- 
tain an integral part and a fractional part. For example, in 
the equation 10 1 - 25 = 17.782 the 1.25 (that is, log 17. 782) 
has 1 for its integral part and 0.25 for its decimal (frac- 
tional) part. In 10 2 = 100 the entire logarithm is integral. 
(Why?) The integral part of a logarithm is called the 
characteristic of the logarithm, and the decimal part is 
called the mantissa of the logarithm. 

The characteristic of a logarithm of any number can 
always be determined at sight. For example: 

log 10 = 1, because 10 1 = 10 ; 

log 100 = 2, because 10 2 = 100 ; 
log 1000 = 3, because 10 3 = 1000 ; 

and so on. But these numbers are all integral powers of 
10. However, the characteristic of the logarithm of any 
other number may be obtained as well by observing what 
powers of 10 inclose it. For example, the characteristic 
of log 525 is 2 because 525 lies between the second and 
the third powers of 10 ; that is, between 10 2 = 100 and 
10 3 = 1000 (see the table, Art. 489). 

It is not so easy to determine the mantissas (the deci- 
mal part) of the logarithms of numbers. We have worked 
out a few of these in the table of Art. 489, but to compute 
the mantissas for all other numbers in this way would be 
a tedious'task. Moreover, the methods necessary to com- 
pute them would be beyond us in difficulty. However, 
these mantissas have been computed for all the various 
powers of 10 (by more advanced methods), and they appear 
in the table of mantissas which follows. So that now when 



LOGARITHMS 429 

we want to know what the logarithm of any number is, 
we decide (by inspection) what the characteristic is and 
then look in the table for the mantissa. 

EXERCISES 

1. Look in the table (pp. 430-431) for the decimal part of 
the logarithms of 10 ; 15 ; 20 ; 38 ; 86 ; "99. 

2. What is the decimal part of the logarithm of 100 ? 

3. What is the power to which 10 must be raised to produce 
10,000 ? 

4. What, then, is the logarithm of 10,000 to the base 10 ? 

5. Examine the table carefully and tell what numbers have 
integers for logarithms ; that is, those that have zero mantissas. 

6. When will a logarithm have a decimal mantissa ? 

7. Find the product of 48 and 55. 
Solution. By means of the table we see that 

48 = 10 1 - 6812 , 
and that 55 = 10 1 - 7404 . 

Therefore 48 x 55 = 10 1 - 6812 x 10 1 - 7404 



The 3 in the exponent 3.4216 tells us that the product of 48 x 55 
is a number between the 3d and 4th powers of 10 ; that is, the 3 
tells us where to put the decimal point. We must find the mantissa 
0.4216 in the table of logarithms and see what number corre- 
sponds to it. 

If we look in the table of mantissas we find that 0.4216 is the 
mantissa of the logarithm of the number 264. Now since the charac- 
teristic of the logarithm is 3, the number must be between the 3d 
and 4th powers of 10 ; that is, between 1000 and 10,000. This means 
that the decimal point comes after the fourth place, so that we must 
add a cipher to 264. Hence the number is 2640. 



430 



GENERAL MATHEMATICS 



TABLE OF MANTISSAS 



No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


10 

11 
12 
13 
14 


0000 
0414 
0792 
1139 
1461 


0043 
0453 
0828 
1173 
1492 


0086 
0492 
0864 
1206 
1523 


0128 
0531 
0899 
1239 

1553 


0170 
0569 
0934 
1271 
1584 


0212 
0607 
0969 
1303 

1614 


0253 
0645 
1004 
1335 
1644 


0294 
0682 
1038 
1367 
1673 


0334 
0719 
1072 
1399 
1703 


0374 
0755 
1106 
1430 
1732 


15 

16 

17 
18 
19 


1761 
2041 
2304 
2553 
2788 


1790 
2068 
2330 
2577 
2810 


1818 
2095' 
2355 
2601 
2833 


1847 
2122 
2380 
2625 
2856 


1875 
2148 
2405 
2648 

2878 


1903 
2175 
2430 
2672 
2900 


1931 
2201 
2455 
2695 
2923 


1959 
2227 
2480 
2718 
2945 


1987 
2253 
2504 
2742 
2967 


2014 
2279 
2529 
2765 
2989 


20 

21 
22 
23 
24 


3010 
3222 
,3424 
3617 
3802 


3032 
3243 
3444 
3636 
3820 


3054 
3263 
3464 
3655 
3838 


3075 
3284 
3483 
3674 
3856 


3(196 
3304 
3502 
3692 

3874 


3118 
3324 
3522 
3-711 
3892 


3139 
3345 
3541 
3729 
3909 


3160 
3365 
3560 

3747 
3927 


3181 
3385 
3579 
3766 
3945 


3201 
3404 
3598 
3784 

3962 


25 

26 
27 
28 
29 


3979 
4150 
4314 
4472 
4624 


3997 
4166 
4330 
4487 
4639 


4014 
4183 
4346 
4502 
4654 


4031 
4200 
4362 
4518 
4669 


4048 
4216 
4378 
4533 
4683 


4065 
4232 
4393 
4548 
4698 


4082 
4249 
4409 
4564 
4713 


4099 
4265 
4425 
4579 
4728 


4116 
4281 
4440 
4594 
4742 


4133 
4298 
4456 
4609 
4757 


30 

31 

32 
33 
34 


4771 
4914 
5051 
5185 
5315 


4786 
4928 
5065 
5198 
5328 


4800 
4942 
5079 
5211 
5340 


4814 
4955 
5092 
5224 
5353 


4829 
4969 
5105 
5237 
5366 


4843 
4983 
5119 
5250 
5378 


4857 
4997 
5132 
5263 
5391 


4871 
5011 
5145 
5276 
5403 


4886 
5024 
5159 
5289 
5416 


4900 
5038 
5172 
5302 

5428 


35 


5441 
5l6T 
5682 
5798 
5911 


5453 
5575 
5694 
5809 
5922 


5465 
5587 
5705 
5821 
5933 


5478 
5599 
5717 
5832 
5944 


5490 
5611 
5729 
5843 
5955 


5502 
5623 
5740 
5855 
5966 


5514 
5635 
5752 
5866 
5977 


5527 
5647 
5763 
5877 
5988 


5539 
5658 
5775 
5888 
5999 


5551 
5670 
5786 
5899 
6010 


"36 
37 
38 
39 


40 

41 

42 
43 
44 


6021 
6128 
6232 
6335 
6435 


6031 
6138 
6243 
6345 

6444 


6042 
6149 
6253 
6355 
6454 


6053 
6160 
6263 
6365 
6464 


6064 
6170 
6274 
6375 

6474 


6075 
6180 
6284 
6385 
6484 


6085 
6191 
6294 
6395 
6493 


6096 
6201 
6304 
6405 
6503 


6107 
6212 
6314 
6415 
6513 


6117 
6222 
6325 
6425 
6522 


45 

46 
47 
48 

49 


6532 
6628 
6721 
6812 
6902 


6542 
6637 
6730 
6821 
6911 


6551 
6646 
6739 
6830 
6920 


6561 
6656 

6749 
6839 

6928 


6571 
6665 
6758 
6848 
6937 


6580 
6675 
6767 
6857 
6946 


6590 
6684 
6776 
6866 
6955 


6599 
6693 
6785 
6875 
6964 


6609 
6702 
6794 
6884 
6972 


6618 
6712 
6803 
6893 
6981 


50 

51 
52 
53 
54 


6990 
7076 
7160 
7243 
7324 


6998 
7084 
7168 
7251 
7332 


7007 
7093 
7177 
7259 
7340 


7016 
7101 
7185 
7267 
7348 


7024 
7110 
7193 
7275 
7356 


7033 
7118 
7202 
7284 
7364 


7042 
7126 
7210 
7292 
7372 


7050 
7135 
7218 
7300 
7380 


7059 
7143 
7226 
7308 
7388 


7067 
7152 
7235 
7316 
73% 


No. 





1 


2 


3 


4. 


5 


6 


7 


8 


9 



LOGARITHMS 



431 



TABLE OF MANTISSAS 



No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


55 

56 
57 
58 
59 


7404 
7482 
7559 
7634 
7709 


7412 
7490 
7566 
7642 
7716 


7419 
7497 
7574 
7649 
7723 


7427 
7505 
7582 
7657 

7731 


7435 
7513 
7589 
7664 

7738 


7443 
7520 
7597 
7672 
7745 


7451 

7528 
7604 
7679 
7752 


7459 
7536 
7612 

7686 
7760 


7466 
7543 
7619 
7694 
7767 


7474 
7551 
7627 
7701 
7774 


60 

61 
62 
63 
64 


7782 
7853 
7924 
7993 
8062 


7789 
7860 
7931 
8000 
8069 


7796 
7868 
7938 
8007 
8075 


7803 
7875 
7945 
8014 
8082 


7810 
7882 
7952 
8021 
8089 


7818 
7889 
7959 
8028 
8096 


7825 
7896 
7966 
8035 
8102 


7832 
7903 
7973 
8041 
8109 


7839 
7910 
7980 
8048 
8116 


7846 
7917 
7987 
8055 
8122 


65 

66 
67 
68 
69 


8129 
8195 
8261 
8325 
8388 


8136 
8202 
8267 
8331 
8395 


8142 
8209 
8274 
8338 
8401 


8149 
8215 
8280 
8344 
8407 


8156 
8222 

8287 
8351 
8414 


8162 
8228 
8293 
8357 
8420 


8169 
8235 
8299 
8363 
8426 


8176 
8241 
8306 
8370 
8432 


8182 
8248 
8312 
8376 
8439 


8189 
8254 
8319 
8382 
8445 


70 

71 
72 
73 
74 


8451 
8513 
8573 
8633 
8692 


8457 
8519 
8579 
8639 
8698 


8463 
8525 
8585 
8645 
8704 


8470 
8531 
8591 
8651 
8710 


8476 
8537 
8597 
8657 
8716 


8482_, 


'8488 
8549 
8609 
8669 
8727 


8494 
8555 
8615 
8675 
8733 


8500 
8561 
8621 
8681 
8739 


8506 
8567 
8627 
8686 
8745 


854~3 
8603 
8663 

8722 


75 

76 

77 
78 
79 


8751 
8808- 
8865 
8921 
8976 


8756 
8814 
8871 
8927 
8982 


8762 
8820 
8876 
8932 
8987 


8768 
8825 
8882 
8938 
8993 


8774 
8831 
8887 
8943 
8998 


8779 
8837 
8893 
8949 
9004 


8785 
8842 
8899 
8954 
9009 


8791 
8848 
8904 
8960 
9015 


8797 
8854 
8910 
8965 
9020 


8802 
8859 
8915 
8971 
9025 


80 

81 
82 
83 

84 


9031 
9085 
9138 
9191 
9243 


9036 
9090 
9143 
9196 

9248 


9042 
9096 
9149 
9201 
9253 


9047 
9101 
9154 
9206 
9258 


9053 
9106 
9159 
9212 
9263 


9058 
9112 
9165 
9217 
9269 


9063 
9117 
9170 
9222 
9274 


9069 
9122 
9175 
9227 
9279 


9074 
9128 
9180 
9232 
9284 


9079 
9133 
9186 
9238 
9289 


85 

86 
87 
88 
89 


9294 
9345 
9395 
9445 
9494 


9299 
9350 
9400 
9450 
9499 


9304 
9355 
9405 
9455 
9504 


9309 
9360 
9410 
9460 
9509 


9315 
9365 
9415 
9465 
9513 


9320 
9370 
9420 
9469 
9518 


9325 
9375 
9425 
9474 
9523 


9330 
9380 
9430 
9479 
9528 


9335 
9385 
9435 
9484 
9533 


9340 
9390 
9440 
9489 
9538 


90 

91 
92 
93 
94 


9542 
9590 
9638 
9685 
9731 


9547 
9595 
9643 
9689 
9736 


9552 
9600 
9647 
9694 

9741 


9557 
9605 
9652 
9699 
9745 


9562 
9609 
9657 
9703 

9750 


9566 
9614 
9661 
9708 
9754 


9571 
9619 
9666 
9713 
9759 


9576 
9624 
9671 
9717 
9763 


9581 
9628 
9675 
9722 
9768 


9586 
9633 
9680 
9727 
9773 


95 

96 
97 
98 
99 


9777 
9823 
9868 
9912 
9956 


9782 
9827 
9872 
9917 
9961 


9786 
9832 
9877 
9921 
9965 


9791 
9836 
9881 
9926 
9969 


9795 
9841 
9886 
9930 
9974 


9800 
9845 
9890 
9934 
9978 


9805 
9850 
9894 
9939 
9983 


9809 
9854 
9899 
9943 
9987 


9814 
9859 
9903 
9948 
9991 


9818 
9863 
9908 
9952 
9996 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 



432 GENERAL MATHEMATICS 

The preceding work may be briefly arranged as follows : 
Let N= 48x55 

log 48 = 1.6812 

log 55 = 1.7404 

Then logA 7 = 3.4216 

By the table, N = 2640. 

493. Logarithm of a product. The discussion and ex- 
amples in Art. 492 have shown that to the problem of 
multiplying two powers of 10 there corresponds the problem 
of adding their logarithms (exponents). This may be stated 
briefly as the first law thus: The logarithm of the product 
of two numbers is the sum of the logarithms of the factors; 
or, by formula, log (aby = log a -f- log b. 

It is easily shown that the law also holds for any number 
of factors in a product ; that is, log abc = log a + log b + log c, 
and so on. 

EXERCISES 

1. Check by actual multiplication the logarithmic method of 
finding the product of 48 and 55. 

2. Find by means of logarithms the products of the follow- 
ing numbers : 

(a) 10 x 100 x 1000. (b) 51 x 40. (c) 83 x 6 x 2. 

3. Find by using logarithms the area of a triangular garden 
plot whose base is 38 ft. and whose altitude is 17 ft. 

Solution. The formula for the area of any triangle is 



Hence, in this case, A = 19 17. 

log 19 = 1.2788 
log 17 = 1.2304 

Then log A = 2.5092 

By the table, A = 323. 



LOGARITHMS 

4. Solve the following problems by means of logarithms : 
4 (a) What is the area of a rectangle whose base is 70 and 
whose altitude is 32 ? 

(b) What is the area of a parallelogram whose base is 64 and 
whose altitude is 85 ? 

(c) What is the area of a square, a side of which is 29 ? 

Solution. The formula for the area of a square is 

A = ,s * = *-. 

In this case .1 = 29 x 29 = 29 2 . 

log 29 = 1.4624 
log 29 = 1.4624 

Therefore log A = 2.9248 

and, by the table, A =841. 

494. Logarithm of a power. Obviously the result in 
Ex. 4, (c), would have been the same if we had multiplied 
the logarithm of 29 by 2 instead of adding it to itself. 
In like manner, if we want the logarithm of 50 3 , we can 
obtain it either by using the logarithm of 50 three times 
as an addend or by taking three times the logarithm of 50. 
Thus, log 50 3 = 3 log 50 = 3 x 1.6990 = 5.0970. 

This discussion illustrates a very important law ; namely, 
that the logarithm of a number roiled to a power in the exponent 
of the power times the logarithm of the number ; or, by for- 
mula, log a" = n log a. 

EXERCISES 

1. Find by the preceding method the logarithms of the 
following numbers: 22 2 ; II 8 ; 15 4 . 

2. Find by logarithms the volume of a cubical cake of ice 
one edge of which is 10 in. ; 11 in. 

3. The area .1 of a circle is given by the formula A = irr, 
where / is the radius of the circle. What is the area of a circle 
whose radius is 6 in. ? (Use logTr = 0.4971.) 



434 GENERAL MATHEMATICS 

495. Logarithm of a quotient. The method of computing 
by logarithms is as useful in division as in multiplication. 
In order to make the method clear, let us review our two 
methods of dividing one number by another. 

(a) - - = 1000, by actual division. 

(b) - = - 2 = 10 3 = 1000, by subtracting the exponents. 

1UU 1U 

Here, as in multiplication, we obtain the same result 
by either method, but the second method reduces the 
operation of actual -division to a simple problem of sub- 
tracting exponents. 

EXERCISES 

1. Find the following quotients by the two methods just 
discussed : 

100,000 1,000,000 

1000 ' 10,000 

2. Divide 100 by 31.623 by using the table of Art. 489. 

So "* ion - = 



XOTE. The student should check this result by actual division. 
3. Divide 562.34 by 31.62 by using the table of Art. 489. 

We may obtain the quotient of any two numbers in 
like manner by subtracting the exponent of the power of 
10 equal to the divisor from the exponent of the power of 
10 equal to the dividend. Keeping in mind the definition 
of a logarithm it is clear that this fact may be expressed as 
a law ; thus, the logarithm of the quotient of two numbers is 
the logarithm of the dividend minus the logarithm of the 

divisor ; or, as a formula, log IT) l9 a ~ 19 ^ 



LOGAEITHMS 435 

EXERCISES 

1. Given log 2 = 0.3010, log 3 = 0.4771, find log f ; log J 

2. Find the value of the following fractions to three 
significant figures by using logarithms : 

x 59 x 85 . 381 x II 3 

< a > 43" ~^~ 

. 752 x 350 71 x 48 x 253 

-~ ~" 



HINT. To find the logarfthm of each fraction, arid the logarithms 
of the factors of the numerator and from this sum subtract the sum 
of the logarithms of the factors of the denominator. 

496. Position of the decimal point. Since the multipli- 
cation or division of a number by 10 amounts to a moving 
of the decimal point one place to the right in multiplication 
and one place to the left in division, and since the multi- 
plication or division of a number by 100 amounts to 
moving the decimal point two places to the right or left, 
and so on, the position of the decimal point in a number 
whose logarithm we are seeking affects the characteristic onli/. 

The truth of the foregoing statement can be seen best by 
means of the table in Art. 489. In this table, for example, 

! O o.25 =1.7782, or log 1.7782 = 0.25. 
If we multiply both sides of this equation by 10, we get 

10 1 - 2B = 17.782, or log 17.782 = 1.25. 

Again, multiplying both sides of this last equation by 10, 
1 02.25 = 177.82, or log 177.82 = 2.25, 

and so on. The student will observe that only the integral 
part of the exponent of 10 (the logarithm of the number) 
is changing, and that the array of figures remains the 



436 GENERAL MATHEMATICS 

same even though the decimal point moves one place to 
the right after each multiplication. In like manner, if we 
divide both sides of the equation 10 - 25 = 1.778 by 10, and 
continue the division, we obtain 

loo.*- 1 = 0.1778, or log 0.1778 = 0.25 - 1; 
100.25-2 = 0.01778, or log 0.01778 = 0.25 - 2, 
lOo.ffi-3 = 0.001778, or log 0.001778 = 0.25 - 3, 

and so on. 

The logarithms 0.25-1, 0.25-2, 0.25-3, etc. are 
negative quantities, but they are not in the form in which 
.we usually write negative numbers. However, if we adopt 
these forms, the mantissas of all our logarithms will not 
only be positive but they will be the same for the same 
array of figures no matter where the decimal point is found. 

Thus the mantissa of log 1.778 is the same as the 
mantissa of log 0.001778, as was shown above. These two 
logarithms differ, therefore, only in their characteristics. 
In some texts log 0.1778 is written 1.25 instead of 0.25 1. 
To agree with this statement the dash above the 1 means 
that only the 1 is negative. Some books prefer the form 
9.25-10 instead of 0.25-1 or 1.25. The student can 
easily see that 9.2510 has the same value as 0.251. 
\Ve shall later see another advantage of the form 9.25 1 0. 

The preceding discussion may be summarized in two 
statements : 

1. We agree to express the logarithm of any number in a 
for tn such that its mantissa shall be positive. This can 
always be done, whether the number is greater or less than 
unity. In either case the positiveness or negativeness of 
the number is shown entirely by means of the characteristic. 

2. Two numbers containing the same succession of digits, 
that ?X differing only in the position of the decimal 



LOGARITHMS 437 

will have logarithms that differ only in the value of the 
characteristic. This explains why we called the table a 
table of mantissas and why, in looking up the logarithm 
of a number, we need pay no attention to the decimal 
point in the number. The same table of mantissas serves 
both for numbers greater and less than unity. 

497. Table of characteristics. The following 'table is 
given in order that the student may determine more 
quickly the characteristic of the logarithm of any number 
between 10 to the minus 6th power and 10 to the plus 
7th power. This is about as much of a range as we 
shall ever need. The table can be extended upward or 
downward at will. 

TABLE OF CHARACTERISTICS 

10~ 6 = 0.000001 10 J =10 

10-5-0.00001 io 2 =ioo 

10- 4 -0.0001 10 3 = 1,000 

io-* = o.ooi io 4 = 10,000 

10-2-0.01.' 10 5 = 100,000 

10- ! = 0.1 io 6 =1,000,000 

10 = 1 IO 7 - 10,000,000 

For example, if we want the logarithm of 2142, we 
know that its characteristic is 3, because 2142 lies between 
the 3d and 4th powers of 10. Again, if we want the 
logarithm of 0.0142, we know that the characteristic of 
the logarithm is 2, because 0.0142 lies between the 
minus 2d and the minus 1st power of 10. 

498. Interpolation. So far we have shown the student 
how to find the logarithms of three-digit numbers only. In 
order to be able to' find the logarithms of numbers of more 
than three digits, and to find the numbers corresponding 



438 GENERAL MATHEMATICS 

to various logarithms which we may obtain in calculation, 
it is necessary for us to learn how to make further use of 
the table in Art. 492. We shall proceed to consider two 
typical examples. 

1. Find the logarithm of 231.6. 

Solution. From the table of Art. 497 it is clear that the char- 
acteristic of the logarithm is 2. To find the mantissa, we ignore the 
decimal point and look in the table of Art. 492 for the mantissa 
of 2316. Reading down the left-hand column of the table, headed 
" No.," we find 23. The numbers in the same horizontal row with 23 
are the mantissas of the logarithms 230, 231, 232, and so on. 

We want to find the logarithm of 2316. We can now write 

log 232 = 2.3655 

log 231 = 2.3636 

Tabular difference = 0.0019 

Now since 231.6 is f' 6 of the way from 231 to 232, we add {'^ 
of the tabular difference 0.0010 to the logarithm of 231. Thus, 

log 231.6 = 2.3636 + ^ x 0.0019. 
Therefore log 231.6 = 2.3647. 

The process of obtaining the logarithm of a number 
in this way is called interpolation. The student should 
practice this method by finding the logarithms of several 
four-digit numbers. 

2. Find the number whose logarithm is 0.3883 1. 

Solution. We know at once that the number is a decimal fraction 
lying between the minus 1st and the power of 10. This tells us 
that the decimal point comes just before the first significant figure 
in the number. If we look in the table of Art. 492 we do not find 
the mantissa 3883, but we find 3892, which is a little greater, and 
3874, which is a little less ; that is, 

0.3892 - 1 = log 0.245. 
0.3874 - 1 = log 0.244. 

Since 0.3883 1 is the logarithm of the number we want, the 
number lies between 0.244 and 0.245. Now 0.3883 - 1 is T 9 8 of the 



LOGARITHMS . 439 

way from log 0.244 to log 0.245 ; hence the number corresponding 
to the logarithm 0.3883 - 1 lies ^, or i, of the way from 0.244 to 
0.245. Therefore the number desired is 0.2445. 

Here the process of interpolation is used on the inverse 
problem, that of finding a number when its logarithm is 
given. 

EXERCISES 

1. Find the logarithms of the following numbers: 745; 83.2; 
91200; 0.567; 0.00741. (No interpolation.) 

2. Find the logarithms of the following numbers : 6542 ; 
783.4; 91243; 0.4826; 0.002143. (Interpolation.) 

3. Find the numbers whose logarithms are 2.6075 : 1.4249; 
0.3054; 0.0212-2; 0.8457-1. (No interpolation.) 

4. Find the numbers whose logarithms are 2.3080; 1.936; 
0.8770 ; 0.0878 - 2. (Interpolation.) 

499. Extraction of roots by means of logarithms. In 
Art. 460 we discussed the meaning of fractional exponents 
and showed that 

a* = Va ; * = Va ; a* = V ; etc. 

If we assume that the theorem of Art. 494, regarding 
the logarithm of a number raised to a power, holds for 
fractional exponents, then 

log Va = log a* = -| log a, 
log v a = log a * = 1 log a, 

and so on. This illustrates the truth of another theorem, 
namely, that the logarithm of any root of a number is equal 
to the logarithm of the number divided by the index of the root. 



9 

Thus, log V542 = J log 542 = ^~- = 1.3670. 

Now 1.3670 is the logarithm of 23.28 -. Therefore the 
square root of 542 is approximately 23.28 . 



440 GENERAL MATHEMATICS 

If the logarithm of the number is negative, the root 
may be found as in the following examples. 
Find by logarithms : 

(a) V0.472. (b) ^0.472. (c) -V/O472. 

Solution. Log 0.472 = 0.6739 1. Now if we attempt to take ^ 
of this negative logarithm, we shall obtain a fractional characteristic 
that would be confusing. Therefore, in order to make it possible 
to keep the mantissa positive and the characteristic (after the 
division) an integer, we write 

log 0,172 = 1!).<>73!> -20, 

a number which the student can readily see is equal to 0.6739 1, 
and which has the added advantage referred to above. We now get 

(a) log Vo.472 .= * (19.6739 - 20) = 9.8369 - 10. 
In like manner, 



(b) log V 0.472 = i (29.6739 - 30) = 9.8913 - 10, 
and () log v 0.472 = J (39.6739 - 40) = 9.9185 - 10. 



In (a), above, log Vo.472 = 9.8369 -10. This means that 
the characteristic is 1 and that the mantissa is 0.8369. 
By reference to the table, we find that 0.8369 is the man- 
tissa of the logarithm of 687. Since the characteristic is 
- 1, the number lies between the minus 1st and the power 
of 10 ; hence the decimal point comes just before the 6, and 
V0.472 = 0.687. The student should check this result by 
actually extracting the square root of 0.472 by the method 
given in Art. 446. 

EXERCISES 

1. Find by logarithms : 

V9604 ; V153.76 ; V0.000529 ; A/10648 ; 
^4275 ; -v/3.375 ; ^0.001728. 

2. Given a = 4.25, ft = 22.1, and c = 0.05, find by logarithms 

I 72 

the value of \| to three significant figures. 



LOGARITHMS 441 



. 182.41 ' , 3)2.42x35.1 

3. Find by logarithms the value of V and -\\ ~^r^ T^ 

i 4.oo^ i i>. X J." 

to three significant hgures. 

4. The velocity v of a body that has fallen a distance of 
,s- feet is given by the formula v = V2 </.->. If g = 32.16, what 
is the velocity acquired by a body in falling 30 ft.? 

5. If a bullet is shot upwards with an initial velocity of v, 
the height .s to which it will rise is given by the equation 
r = V2 gs. What must be the minimum velocity at the start 
of a bullet fired upwards if it is to strike a Zeppelin 1500 ft. 
high? (Assume y = 32.16.) 

6. The time t of oscillation of a pendulum / centimeters long 

is given by the formula t = 7r\| j- Kind the time of oscilla- 

i j)oO 

tion of a pendulum 78.22 centimeters long. 

* 7. If the time of oscillation of a pendulum is 1 sec., what 
is its length ? 

*8. The area of any triangle is given by the formula 

A = Vs (s a) (s b} (s c), 

where a, b, and c are the sides of the triangle, and s equals 

a + b 4- c 

one halt the perimeter, or 

& 

Using the preceding formula, find by means of logarithms the 
number of acres in a triangular field whose sides are 15, 38.2, 
and 45.3 rods respectively. (1 A. = 160 sq. rd.) 

*9. The area A of the cross section of a chimney in square 
feet required to carry off the smoke is given by the formula 

0.6 P 

/~ ' 

V// 

where P is the number of pounds of coal burned per hour, and 
h is the height of the chimney in feet. Find out what the area 
of a cross section of a chimney 80 ft. high should be to carry 
off the smoke if 560 Ib. of coal are burned per hour. 



442 



GENERAL MATHEMATICS 



* 10. The average velocity v of the piston head in inches per 
second for a steam engine is given by the formula 



where s denotes the distance (in inches) over which the piston 
head moves, and p is the number of pounds of pressure of 
steam in the cylinder. Find v if s = 30.24 in. and p = 115 Ib. 

*11. In the equation x W 1J what is the value of x when 
y = ? when y 1 ? when y = 2 ? when ?/= !? 

*12. Fill in the following table for the equation x =10 ?/ . 



y 





i 


2 


-1 


- 2 


-3 


i 


I 


i 


4 


5 


X 


i 


10 


100 



















NOTE. For y equaling ^, ^, 2^, etc., use the table of mantissas, 
Art. 492. 

* 13. Plot the results in the table of Ex. 12 and draw the curve, 
thus obtaining the'graph for x = lO 2 ' (or y log x) (Fig. 300). 




FIG. 300. GRAPH OF x = 



*14. Show that the graph for x =10^ (Fig. 300) makes clear 
the following principles : 

(a) A negative number does not have a real number for its 
logarithm. 



LOGARITHMS 443 

(b) The logarithm of a positive number is positive or nega- 
tive according as the number is greater or less than 1. 

(c) The greater the value of x, the greater its logarithm. 

(d) As x gets smaller and smaller, its logarithm decreases 
and becomes smaller and smaller. 

*15. Find by the graph of Ex. 13 the logarithm of 2.25; 
of 4.5 ; of 1.1 ; of 2.8. 

*J6. Of what number is 0.35 the logarithm? 0.5? 0.42? 
*17. Check your results for Exs. 15 and 16 by the results 
given in the table of Art. 492. 

* 500. Exponential equations. Instead of finding the log- 
arithm of 1000 to the base 10, we could arrive at the same 
result by solving the equation 10 r = 1000, for this equa- 
tion asks the question, What power of 10 equals 1000? 
In other words, What is the logarithm of 1000 to the 
base 10 ? An equation like this, in which an unknown is 
involved in the exponent, is called an exponential equation. 

EXERCISE 

Give five examples of exponential equations where 10 is 
the base. 

*501. Method of solving exponential equations. The sim- 
plest exponential equations may be solved by inspection 
just as the logarithms of many numbers can be given by 
inspection. Where an exponential equation cannot be 
solved readily by inspection, logarithms may be employed 
to simplify the process. We will illustrate each case. 

Solution I (by inspection). 

(a) If 2 X = 4, then x = 2. (e) If 10?' = 100, then y - 2. 

(b)If3 a; = 9, then* = 2. (f)IflO j; = 1000, then x - 3. 

(c) If 2V = 8, then / = 3. (g) If 10 = 10,000, then y = 4. 

(,1) Tf 3* = 81, then x = 4. 



444 GENERAL MATHEMATK \S 

Solution II (by using logarithms). 

Solve the equation 2 X = 6 for x. 
Taking the logarithms of both sides : 

log 2 X = log 6, 
or x log 2 = log 6. 

. z = !2e = 02 = 2 58 
log 2 0.3010 

1 G. 

The student must remember that ^ - is not equal to 

6 log2 

log - The first is a fraction obtained by dividing one 

A 

logarithm by another, and involves division ; the second 
indicates that the logarithm of a fraction is to be found, 
and involves subtraction. 

EXERCISE 

Solve the following equations for x : 

(a) 2 X = 7. (b) 3 X = 5. (c) 4* = 10. (d) (1.12)* = 3. 

502. Interest problems solved by logarithms. Some im- 
portant problems in interest may be solved by means of 
exponential equations and logarithms. The following 
simple example will illustrate the principle: 

In how many years will a sum of money double itself at 6% 
if the interest is compounded annually ? 

Solution. In one year $1 will amount to $1.06 ; in two years the 
amount will be 1.06 x 1.06, or (1.06) 2 ; in three years the amount 
will be (1.06) 3 ; and so on. Therefore in x years the amount of $1 
will be (1.06) 3 '. Then, if the money is to double itself in x years, 
the conditions of the problem will be represented by the equation 
(1.06)'* = 2. Solving this equation for x, we get x = 12.3 (approx.). 
Therefore a sum of money will double itself at 6% (compounded 
annually) in about 12.3 yr. 



LOGARITHMS 445 

EXERCISES 

1. Explain the solution of the problem given in Art. 502. 

2. If the interest is compounded annually, in how many years 
will a sum of money double itself at 3% ? 3|-% ? 4% ? 5% ? 

3. In how many years will a sum of money treble itself at 
4^ interest compounded annually ? semiannually ? 

4. The amount of P dollars for n years at /%, compounded 
annually, is given by the formula .1 =P(1 +/)". Find the 
amount of $1200 for 10 yr. at 4%. 

Solution. In this problem P = 1200, r = 0.04, n = 10. 

^ = 1200(1 + 0.04) 
The computation may be arranged as follows : 

log 1200 = 3.0792 

10 log 1.04 = 0.1700 

log A = 3.2492 

Therefore A = 1775, number of dollars in the 

amount. 

NOTE. As a matter of fact, this value of A is not exact, because 
we are using only four-place tables. In practice the value of the 
problem should determine the kind of tables used. The greater the 
number of places given in tin' tables used, the greater the accuracy 
of the result. 

5. What will $5000 amount to in 5 yr. at 3%, interest 
compounded annually ? semiannually? quarterly? 

6. Approximately three hundred years ago the Dutch 
purchased Manhattan Island from the Indians for $24. What 
would this $24 amount to at the present time if it had been 
placed on interest at 6%' and compounded annually? 

7. What would he the amount of 10 at the present time 
if it had been placed on 3% annual compound interest tifty 
years ago ? 



446 GENERAL MATHEMATICS 

8. A boy deposited 300 in a savings bank on 3% interest, 
the interest to be compounded annually. He forgot about his 
deposit until fifteen years later, when he found the receipt 
covering the original deposit. What did the 300 amount to in 
the fifteen years ? 

9. What sum will amount to $1600 in 10 yr. at 6%, interest 
being compounded annually ? 

10. What sum will amount to $ 2500 in 5 yr. at 3%, interest 
being compounded annually ? 

11. In how many years will $4000 amount to $8500 at 6%, 
interest being compounded annually ? 

12. What would be the amount to-day of 1 cent which 
nineteen hundred and twenty years ago was placed on interest 
at 6%, compounded annually ? Find the radius in miles of a 
sphere of gold which has this value. 

NOTE. A cubic foot of gold weighs 1206 pounds avoirdupois, 
one pound being worth approximately $290. 

The volume of a sphere is given by the formula V = 3 irr 3 , where 
V is the volume and r the radius of the sphere. 

No doubt the pupil is convinced of the value of log- 
arithms as a labor-saving device in complicated arithmetic 
computations. Since he will meet numerous opportunities 
for applications, the lists of problems in the chapter are 
brief, the aim being to give just enough illustrations to 
make clear the meaning of the principles involved. 

HISTORICAL NOTE. Logarithms were invented by John Napier 
(1550-1617), baron of Merchiston in Scotland. His greatest purpose 
in studying mathematics was to simplify and systematize arithmetic, 
algebra, and trigonometry. The student should read about Xapier's 
"rods," or "bones," which he designed to simplify multiplication and 
division (Encyclopaedia Britannica, llth ed.). 

It was his earnest desire to simplify the processes that led him 
to invent logarithms ; and, strange as it may seem, he did not consider 
a logarithm as an exponent. In his time the theory of exponents was 



LOGARITHMS 447 

not known. A Swiss by the name of Jobst Biirgi (1552-1632) may 
have conceived the idea of logarithms as early or earlier than 
Napier and quite independently of him, but he neglected to pub- 
lish his results until after Napier's logarithms were known all over 
Europe. 

Henry Briggs (1561-1630), who, in Napier's time, was professor of 
geometry in Gresham College, London, became very much interested 
in Napier's work and paid him a visit. It is related that upon 
Briggs's arrival he and Napier stood speechless, observing each other 
for almost a quarter of an hour. At last Briggs spoke as follows : 
" My lord, I have undertaken this long journey purposely to see your 
person, and to know by what engine of wit or ingenuity you came 
first to think of this most excellent help in astronomy, namely, the 
logarithms, but, my lord, being by you found out, I w y onder nobody 
found it out before, when now known it is so easy." 

After this visit Briggs and Napier both seem to have seen the 
usefulness of a table of logarithms to the base 10, and Briggs devoted 
himself to the construction of such tables. For this reason logarithms 
to the base 10 are often called Briggsian logarithms. 

Abbott says that when Napoleon had a few moments for diversion, 
he often spent them over a book of logarithms, which he always 
found recreational. 

Miller in* his " Historical Introduction to Mathematical Litera- 
ture (p. 70) says : "The fact that these logarithms had to be computed 
only once for all time explains their great value to the intellectual 
world. It would be difficult to estimate the enormous amount of time 
saved by astronomers and others through the use of logarithm tables 
alone." (For further reading see Cajori's "History of Elementary 
Mathematics," pp. 155-167. Consult also the New International Cy- 
clopedia, which contains a great deal of excellent historical material.) 

SUMMARY 

503. This chapter has taught the meaning of the follow- 
ing words and phrases : logarithm, characteristic, mantissa, 
interpolation, and exponential equation. 

504. The theory and practical value of logarithms has 
been discussed in as elementary a way as possible so that 



448 GEN KKA L MATHEMATK 'S 

the student may be able to appreciate the value of this 
powerful labor-saving device. 

505. This chapter has taught the student four impor- 
tant logarithmic formulas : 

(a) log ab log a + log b* (c) log a u = n log a. 

(b) log - = log a log b. (d) log Va = 

o n 

506. The position of the decimal point in any result 
depends entirely upon the characteristic of the logarithm 
of the number sought. Thus, two numbers having the 
same succession of digits will have the same mantissa. 
The mantissa of the logarithm of a number is always 
positive; the characteristic may be either + or . 

507. This chapter has taught methods of solving loga- 
rithmic and exponential equations. 

508. The student has been taught how to solve verbal 
problems by means of logarithms, for example, the interest 
problem. 



CHAPTER XVIII 

THE SLIDE RULE AND GEOMETRIC REPRESENTATION OF 
LOGARITHMS PRACTICALLY APPLIED TO THE SLIDE RULE 

509. General description of the slide rule. The theory of 
the slide rule is based on the elementary ideas and prin- 
ciples of logarithms, and anyone can learn to use the slide 
rule who has the ability to read a graduated scale. 

The slide rule may be used in nearly all forms of cal- 
culation and is gradually coming into general use in 
many different fields: to the practicing engineer and to 
the student of the mathematical sciences it is invaluable ; 
to the accountant and the statistician it is an instru- 
ment of great service. The student will therefore find it 
advisable to learn all he can about the actual use of this 
important device. 

In Chapter II we discussed the graphical method of 
adding and subtracting line segments as an interpretation 
of the addition and subtraction of numbers. Since to a 
multiplication of two numbers there corresponds an addi- 
tion of their logarithms, it is possible to obtain the loga- 
rithm of a product graphically by adding line segments 
whose lengths represent the logarithms of the factors. In 
order to carry out this plan, a method of actually finding 
a line segment whose length shall represent the logarithm 
of a given number has been developed. 

The slide rule is a mechanical device for determining 
products, quotients, powers, and roots of numbers by 
graphical addition and subtraction. 

449 



450 



GENERAL MATHEMATICS 



Mantissas of logarithms of numbers from 1 to 10 (which, 
as we have seen, are the same for numbers from 100 to 1000 
or from 1000 to 10,000) are laid off to a certain scale on 
two pieces of rule (see Fig. 301) which are made to slide by 



nr 




FIG. 301. SLIDE KUL.K 

each other so tliat the sums or differences of the logarithms 
can be obtained mechanically. 

The scale is numbered from 1 to 10 at the points which 
mark the logarithms of the several numbers used. The com- 
mon scale is 5 in. long and the common rule 10 in. long, so 
that the series of logarithms is put on twice, and the number- 
ing either repeated for the second set or continued from 10 



A 1 


2 34 68 






B 1 2 34 
C 




D 







FIG. 302 

to 100. The most common form of the rule is the Mannheim 
Slide Rule. On this rule, which is made essentially as shown 
in Figs. 302 and 303, there are two scales A and B just alike 
(A on the rule and B on the slide), and two other scales, 
C and D, just alike (Z> on the rule and C on the slide). 

The student will note that the distance from 1 to 2 on 
scales A and B is the same as the distance from 2 to 4 and 



451 



from 4 to 8. This means that if we add the distance from 
1 to 2 on scale B to the distance from 1 to 2 on scale A by 
means of the slide B, we shall obtain the product 2x2, 
or 4. In like manner, if we add the distance from 1 to 
4 on scale B to the distance from 1 to 2 on scale A, we 
shall obtain the product 4 x 2, or 8. 

C and D differ from A and B in being graduated to a 
unit twice as large as the unit to which A and B are 
graduated, so that the length representing the logarithm of 
a given number on C and D is twice as long as the length 



A 1 



D 



FIG. 303 

on A and B representing the. logarithm of the same number. 
Therefore, any number on the lower rule or slide is oppo- 
site its square on the upper rule or slide ; and if the upper 
rule and slide be regarded as standard logarithmic scales, 
the lower rule and slide will be standard logarithmic scales 
of the squares of the numbers shown. The student can 
easily understand from the preceding statement how the 
square roots of numbers are found. 

It should be said, however, that the values of the loga- 
rithms themselves are not shown on the scales. What we 
find is the numbers which correspond to the logarithms. 
Each unit length on the scales (graduated lengths) repre- 
sents equal parts of the logarithmic table. Thus, if the 
logarithm of 10 be selected as the unit, then the logarithm 



452 



of 3, or 0.477, will be represented by 0.477 of that unit; 
4 by 0.602 ; 5 by 0.699 ; and so on, as can be seen by 
referring to the following table of corresponding values : 



Number 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


Logarithm 





0.301 


0.477 


0.602 


0.699 


0.778 


0.845 


0.903 


0.954 


1.000 



The numbers between 1 and 2, 2 and 3, 3 and 4, and 
so on, are represented on the logarithmic scales by inter- 
mediate divisions, and the entire scale has been graduated 
as closely as is possible for convenience in reading. 

The preceding discussion should therefore make it clear 
that at the ^y^th division along the scale on the slide rule 
we should find 2 and not its logarithm, and at the ^nnj^h 
division we should find 5 and not 0.699, and so on. 

It is clear that this scheme eliminates entirely the 
process of finding the numbers corresponding to certain 
logarithms, as we had to do when we computed by means 
of the table in Art. 492. 

The student will observe, further, that the left index of 
the scales (that is, the division marked 1) may assume 
any value which is a multiple or a decimal part of 1 (for 
example, 10, 100, 1000, 0.1, 0.01, 0.001, etc.), but when 
these values are assumed, this same ratio must be held 
throughout the entire scale. In this case the proper values 
for the subsequent divisions of the scale in order would be 
20, 30 ; 200, 300 ; 2000, 3000 ; 0.2, 0.3 ; 0.02, 0.03 ; 0.002, 
0.003. It follows that as the value of the 1 at the begin- 
ning of each scale varies any number such as 382 may 
have the value 38200, 3820, 382, 38.2, 3.82, 0.382, 0.0382, 
and so on. 

If a number which the student has to read does not 
come exactly at a graduation he must estimate the values 



THE SLIDE RULE 453 

as closely as possible ; for example, if a certain number were 
indicated ^ of the way from 152 to 153, he would read 
152.3, on the assumption, of course, that the left index of 
the scale has the value 100. We shall now proceed to 
show by specific examples how the slide rule is used. 

510. Multiplication with the slide rule. All calculations 
in multiplication, division, and proportion are worked out 
on scales C and Z>, as by reason of the greater space allotted 
to each of the divisions the results obtained are more 
accurate. To find the product 2x3 set scale C so that 
its left-hand index (the division marked 1) falls exactly 
opposite the division marked 2 on scale D (see Fig. 302). 
Then directly opposite the division marked 3 on scale C 
we shall find on scale D the division marked 6, which is 
the required product. 

This process is justified by the fact that to log 2 on 
scale D we add log 3 on scale (7, thus obtaining log 6 
on scale D. 

In general, to multiply any constant number a by another 
number ft, set 1 of scale C opposite a of scale D and read the 
product ab on scale D opposite b of scale C. 

EXERCISES 

1. Use the slide rule to find the products in the following 
problems : 

(a) 2 and 4 ; 2 and 5 ; 2 and 6 ; 2 and 7 ; 2 and 8 ; 2 and 9 ; 

2 and 10. 

(b) 3 and 2 ; 3 and 3 ; 3 and 4 ; 3 and 5 ; 3 and 6 ; 3 and 7 ; 

3 and 8 ; 3 and 9 ; 3 and 10. 

2. How would you find the product of 20 and 30 ? 

NOTE. The Mannheim Slide Rule will enable us to secure results 
correct to three significant figures, and in exceptional cases results 



454 GENERAL MATHEMATICS 

correct to even four significant figures may be obtained. However, 
in the work of this chapter we shall be content if we make our com- 
putation correct to two, or perhaps three, significant figures, because 
in actual practice this is sufficient. 

511. Division with the slide rule. To divide 6 by 3, set 
3 of scale C opposite 6 of scale D (see Fig. 302) and read 
the quotient 2 on scale D opposite 1 of scale C. This 
process is justified from the fact that from log 6 we sub- 
tract log 3, thus obtaining log 2 on scale D. 

In general, to divide any constant number a by another 
number 6, set b of scale C opposite a of scale D and read the 

quotient - on scale D opposite 1 of scale C. 

EXERCISE 

1. Use the slide rule to find the quotients in the following 
problems : 

(a\ 4.6-8.10 /Vl 8 . 8 8 . 8 ( n\ 10- 10. 10- 10 
W 2' 2' 2> ~2~" W f f ' * f* \- ) "3"' ~4~' ~~' TO* 

512. The runner. Each slide rule is equipped with a 
runner which slides along the rule in a groove and by 
means of which the student is enabled to find more quickly 
coincident points on the scales. It is also valuable to mark 
the result of some part of a problem which contains several 
computations. Thus, if 

25.2 x 3.5 x 3.68 
-22*- 

we can compute 25.2 x 3.5 by one setting of the slide and 
then bring the runner to 88.2, the result. We then bring 
the index of the slide to 88.2 and multiply by 3.68 ; this 
gives 324.6 (approx.). Bring the runner to this result 
and divide by 22.6; the quotient should be 14.4 (approx.). 



THE SLIDE RULE 455 

The student can easily determine the use of the runner 
in finding powers and roots after he has read the next 
two articles. 

513. Raising to powers with the slide rule. The student 
will observe that if the logarithm of any number in the 
table of Art. 492 be multiplied by 2, the logarithm thus 
obtained will correspond to the square of that number. 
Thus, if the logarithm of 2 (0.301) be multiplied by 2, the 
result (0.602) is the logarithm of 2 2 , or 4. This is in 
accord with the law of Art. 494 regarding the logarithm 
of a number raised to a power. 

In like manner, the logarithm of 2 multiplied by 3 is 0.903, 
which is the logarithm of 2 3 , or 8, and so on. Since this same 
relation holds for any number to any power, we may raise 
a number to any power by using the slide rule as follows : 

1. Squares of numbers. 

To find the value of S 2 look for 3 on scale D and read 
3? = 9 directly opposite 3 on scale A. 

2. Cubes of numbers. 

To find the value of 3 s , set 1 of scale B opposite 3 of 
scale A and find 3 s , or 27, on scale A opposite 3 of scale C. 

3. Fourth power of numbers. 

To find the value of 3*, set 1 of scale C to 3 on scale D 
and find 3*, or 81, on scale A opposite 3 of scale C. 

4. Higher powers. 

Higher powers of numbers may be found by a method 
similar to the preceding, but we shall not need to discuss 
these here, as most of our problems will deal with the lower 
process of numbers. 

514. Square roots found by means of the slide rule. To 
find the square root of any number, bring the runner to 
the number on scale A, and its square root will be found 



456 GENERAL MATHEMATICS 

at the runner on scale D exactly opposite the number. 
This process is seen to be the exact inverse of finding the 
square of a number. 

If the number contains an odd number of digits to the 
left of the decimal point, its square root will be found on 
the left half of the rule ; if it contains an even number 
its square root will be found on the right half. If the 
number is a fraction, and contains an odd number of zeros 
to the right of the decimal point, the root is on the left 
half ; if it contains an even number (or no zeros) the root 
is on the right half. If the student prefers he may deter- 
mine the first figure of the root mentally and then find 
the proper half of the rule to use by inspection. 

EXERCISES 

1. Find the square roots of the following numbers by means 
of the slide rule and compare your results with those of the 
table in Art. 449 : 169 ; 576 ; 625 ; 900 ; 2.25 ; 3.24 ; 1.96 ; 4.41. 

2. What is the side of a square whose area is 784 sq. ft.? 

515. Cube roots found by means of the slide rule. To 
find the cube root of a number by means of the slide rule, 
move the slide either from left to right or from right to 
left until the same number appears on scale B opposite the 
given number on scale A as appears on scale D opposite the 
left or right index (division 1) on C. The number which 
appears on both scales B and D is the cube root of the 
given number. For example, to find the cube root of 125, 
move the slide to the left till 5 on scale B appears opposite 
125 of scale A, and 5 on scale D lies opposite 1 (the right 
index) on scale C. Thus 5 is the cube root of 125. 

A second method of finding the cube root is to invert 
the slide (see Art. 517) and set 1 on scale C under the 



THE SLIDE RULE 457 

given number on scale A and then find the number on 
scale B which lies opposite the same number on scale D. 
This number will be the cube root sought. The student 
should observe that the process of extracting the cube 
root is the inverse of that of cubing a number. 

EXERCISES 

1. Find the cube roots of the following numbers by means 
of the slide rule and compare your results with those of the 
table in Art. 449 : 64 ; 125 ; 0.729 ; 2197 ; 0.001331. 

2. Find one edge of a cube whose volume is 1728 cu. in. 

516. Proportion. A great many problems in proportion 
can be solved very easily by means of the slide rule. 
The student will observe that if the left-hand indexes 
of scales C and D coincide, the readings on the slide 
are in direct proportion 1:1 to the opposite readings on 
the rule. 

In like manner, if the left index of C is made to coin- 
cide with 2 on D, the ratio will be 1 : 2, and so on. Hence, 

10 25 

to find the fourth term in the proportion = , we 

15 x 

move the slide to the right until 10 on scale C is opposite 
15 on scale D, and opposite 25 on C read off x = 37.5 on D. 
In general, to find the fourth term in a proportion, set the 
first term of the proportion on scale C to the second term on 
scale D and find the fourth term (unknown) on scale D oppo- 
site the third term on C. 

517. Inverted slide. The slide may be inverted so that 
scale C faces scale A. This gives inverted readings on 
scale C which are the reciprocals of their coincident read- 
ings on scale D, and vice versa. Thus, 3 on scale D lies 
opposite 0.33 on scale (7, and vice versa. 



458 GENERAL MATHEMATICS 

The inverted slide is useful, as is shown in finding cube 
roots and also in problems involving an inverse propor- 
tion, as in the following example : 

If 10 pipes can empty a cistern in 22 hr., how long will it 
take 50 pipes to empty the cistern ? 

Solution. Invert the slide and set 10 of scale B at 22 on scale D, 
and opposite 5 on C find 4.4 (the result) on scale D. 

518. Position of the decimal point. The student will be 
able in most practical problems to determine the position 
of the decimal point. If there is any considerable difficulty 
in any later work he should consult some standard manual 
on the use of the slide rule. 

MISCELLANEOUS EXERCISES 
Solve the following problems by means of the slide rule : 

1. Find the product of 58.2 x 2.55; 33.4 x 75.6; 22.5 x 
33.3 x 8.2 ; 0.12 x 0.09 x 0.003. 

82.5 3.5 0.04 

2. tind the following quotients: ; ; ; ; 

35.3 x 75.5 . 7.2 x 83.5 x 0.09 
22.8 3.6 

3. Perform the indicated operations: 25 2 ; 3 3 3 ; 2 x 2 2 x 15 8 ; 
12 4 ; 1.2 2 x 7.5 2 x 0.9 2 . 

4. Extract the indicated roots: V2? V3; V5; V7; Vl2; 
V576; VTO6; Vl37.2 ; -^8; -^15; -\/Tl2; A/64; ^1728. 

5. Find the circumference of a circle whose diameter is 6; 
5.5 ; 2.83. (See a slide-rule manual for a short cut.) 

6. If 10 men can do a piece of work in 4 da., how long 
will it take 5 men to do the work if they work at the 
same rate? 

7. What will be the cost of 13| ft. of rope at 3j$ per foot ? 



THE SLIDE RULE 459 

8. What is the volume of a cubical stone 6.3 ft. long, 4.5 ft. 
wide, and 3.2 ft. high ? 

9. What distance will a train travel in 10 hr. and 20 min. 
at a rate of 30.5 mi. per hour ? 

10. What is the interest on $5600 for 1 yr. at 3J% ? 

at 5%? at 6%? 

11. In how many hours will a train travel 756 mi/ if it 
travels at the rate of 41.2 mi. per hour ? 

12. How many miles in 2783 ft. ? in 17,822 ft. ? 

13. Find the area of a circle whose diameter is 10 in.; 
7.5 in.; 0.351 in. (Use A = 0.7854 d 2 .) 

14. Find the volume of a cube one of whose edges is 
3.57 in. ; of a cube one of whose edges is 82.1 in. 

15. If a man can save $3120 in 26 mo., how many dollars 
will he save at the same rate in 1 mo. ? 6 mo. ? 12 mq. ? 

16. Find the mean proportional between 6 and 27. 

17. If goods cost 550 a yard, at what price must they be 
sold to realize 15% profit on the selling price ? 

18. The formula for the area of the ring in Fig. 304 is 
A = irR* Trr 2 . Since 2 r = d, this formula may be 

written A- =- 1, or A= ~^J 2 ~ 

Using the last formula, find the area of the ring 

when R = 8.5 in. and r =' 5.3 in. FIG. 304 

19. Find the amount of $225 invested for 12 yr. at 6% 
simple interest. (Use the formula found under logarithms or 
refer to a slide-rule manual for a short cut.) 

20. What force must be applied to a lever 5.2 ft. from the 
fulcrum to raise a weight of 742 Ib. which is 1| ft. on the 
opposite side of the fulcrum ? 




460 



GEXK 1 1 A L M AT H KM ATICS 




21. A shaft makes 28 revolutions and is to drive another shaft 
which should make 42 revolutions. The distance between their 
centers is 60 in. What should be the 

diameter of the gears ? (See Fig. 305.) 

HINT. Refer to a slide-rule manual 
or, better, develop the formula between 
the diameter of the gears, the number 
of revolutions, and the distance between 
their centers, as follows : FIG. 305 

28 circumferences of the large gear = 42 circumferences of the small 
gear. Why '.' 

Then 28 (60 - x) = 42 x (see Fig. 305). 

28 60 = 70 x. Why ? 

| = f. 

Then apply the slide rule. 

22. Since F = ^C + 32 is the equation representing the 
relation between the Fahrenheit and centigrade thermometers, 
we may compare readings of these two thermometers by the 
following scheme : 



c 


Set 5 


Below degrees centigrade 


I) 


to 9 


Read degrees plus 32 equals Fahrenheit 



W r hat then is F. when C. = 25 ? 18 ? 

*23. Trigonometric applications are greatly simplified by the 
slide rule. The solution of a formula 

be sin^4 . . 

like A = is readily obtained. 

Sin A may be used directly as a factor 
in performing the operation. Find the 
area of the corner lot in Fig. 306. 




6=84' 

FIG. 306 

*24. Find the value of the lot in Ex. 23 at $871.20 per acre. 

NOTE. For numerous applications of the slide rule to practical 
problems the student should consult a standard slide-rule manual. 



THE SLIDE KULE 461 

SUMMARY 

519. This chapter has taught the meaning of the follow- 
ing words and phrases: slide rule, runner, inverted slide. 

520. The theory and practical value of the slide rule 
have been discussed and illustrated so that the student 
can get at least an elementary working knowledge of this 
powerful labor-saving device. 

521. The student has been taught how to use the slide 
rule in multiplying, dividing, raising to powers, and ex- 
tracting roots. 

522. The student has been shown how problems in 
proportion and many other verbal problems may be solved 
by the slide rule. 

523. The student has been referred to the slide-rule 
manual for methods of solving the more difficult problems. 



CHAPTER XIX 

QUADRATICS; QUADRATIC FUNCTIONS; QUADRATIC EQUA- 
TIONS; GRAPHS OF QUADRATIC EQUATIONS; FORMULAS 
INVOLVING QUADRATIC TERMS 

524. Quadratic-equation problem. An engineer increased 
the speed of his train 2 mi. an hour and made a run of 
180 mi. in 1 hr. less than schedule time. What was the 
speed when running according to schedule ? 

Solution. Let x = the ordinary rate of the train. 
Then x + 2 = the rate after the increase. 

180 

= the schedule time. 
x 

180 

= the time it takes after the speed is increased. 



x + 2 

180 180 

Then = hi. A\ hv . ; 

x x + 2 

The L.C.D. is x (x + 2). Why ? 

Multiplying through by x (x + 2) we get 
180 (x + 2) = ISOx + x(x + 2). 
180 x + 360 = 180 x + x 2 + 2 x. 

360 = x* + 2x. Why ? 

x 2 + 2x- 360 = 0. Why? 



We are not able to simplify the equation o?+ 2# 360 = 
further. The methods of the preceding chapters will not 
solve it. In fact we appear to have come to the end of 
the road. It is clear that the problem is solved if we 
can find a value of x which will make the value of the 
quadratic trinomial z 2 + 2 x 360 equal to zero. 

462 



QUADRATIC EQUATIONS 



463 



An equation in which the highest power of the unknown 
is the second power is called a quadratic equation. Many 
problems in geometry, science, and mechanics are solved 
by quadratic equations. It is our purpose in this chapter 
to develop the power to solve quadratic equations and to 
apply quadratic methods to verbal problems. This process 
will be illustrated by the solution of the given equation, 
3? + 2 x - 360 = 0. 

525. Quadratic function. The expression x* + 2 x 360 
is a quadratic function of x, or a function of the second 
degree ; with every change in the value of x the value of 
the function x 2 + 2 x 360 changes. We shall get some 
light on the question, What value of x will make the ex- 
pression 3? + 2x 360 equal to ? by studying how the 
value of the expression x 2 + 2 x 360 changes as we give 
different values to x. This variation is best shown by 
means of the graph. 

526. Graph of a quadratic function. In order to under- 
stand more about the graph of a quadratic function we 
shall consider a few simple exercises. 

INTRODUCTORY EXERCISES 

1. Find the value of the function a- 2 + 2 a- 360 for each of 
the following values of x : 0, 10, 10, - 15, 15, 20, 18, 19, 21. 

2. Fill in the following table of corresponding values 
for x and the function x 2 + 2 x 360 : 



X 





10 


-10 


+ 15 


-15 


+ 16 


-16 


+ 17 


-17 


J 2 + 2x-360 


-360 


-240 


-280 
















x 


+ 19 


-19 


+ 21 


-21 


+ 25 


-25 


+ 30 


-30 


x 2 + 2x-360 



















464 



GENERAL MATHEMATICS 



If we transfer the corresponding values of x and the function 
x 2 + 2 x 360 from the table to a sheet of cross-section paper 







































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FIG. 307. SHOWING THE GRAPH OF A QUADRATIC FUNCTION 

so as to secure the points which correspond to these values, 
we shall obtain a series of points which, when connected by 
a smooth curve, will be like the curve in Fig. 307. 

EXERCISES 

1. From the graph (Fig. 307) determine how the value of 
a-2 _j_ 2 x _ 360 changes as x changes from 25 to 20 ; from to 
15 ; from to - 15 ; from - 25 to - 30 ; from + 25 to + 30. 

2. What value of x will make the function or 2 + 2x 360 
equal to 300? 200? 150? -250? -300? 

The preceding exercises show us that the graph enables 
us to see what value x must have in order that the expres- 
sion a^+ 2 x 360 shall have a given value. The pupil will 
recall that, in the speed problem with which we started, the 



QUADRATIC EQUATIONS 465 

one question we could not answer was, What value of x 
will make the expression a^+ 2 x 360 equal to zero ? This 
question is now easily answered. A glance at the graph 
shows that the expression becomes at two places ; namely, 
when a: = 18 or when a; = 20. Thus the equation a?+2x 
360 = is satisfied by x = 18 or x = 20. Check these 
values by substituting in the equation. Hence the speed 
of the train running according to schedule was 18 mi. per 
hour. We reject the 20 as meaningless. (Why ?) 

527. The two solutions of a quadratic equation. In the 
preceding article we rejected 20 as a meaningless solution. 
Though a quadratic equation has two solutions, this does 
not mean that every verbal problem that leads to a quad- 
ratic equation has two solutions. The nature of the con- 
ditions of the verbal problem may be such as to make one 
or even both of the solutions of the quadratic impossible 
or meaningless. When neither of the two solutions of the 
quadratic is a solution of the problem it usually means 
that the conditions of the problem are impossible, or that 
the problem is erroneously stated. In fact it would be 
easy to make up an arithmetic problem whose result 
could not be interpreted ; for example, it would be diffi- 
cult to jump out of the window 2i times. To decide 
which solution, if either, meets the conditions stated in 
a verbal problem, it is necessary to reread the problem, 
substituting the solutions in the conditions of the prob- 
lem, and to reject solutions of the equation which do 
not meet the conditions. 

528. How to solve a quadratic equation graphically. We 
may now summarize the method of solving a quadratic 
equation revealed in the discussion of Arts. 526-527 as fol- 
lows : (1) Reduce the equation to the form ax 2 + bx + c = 0. 
(2) Make a table showing the corresponding values of x and 



460 



G EN ER A L M AT 1 1 E M ATICS 



the function ax 2 -f bx + c. (3) Transfer the data of the 
table to squared paper and construct the curve represent- 
ing the function ax 2 + bx + c. This curve shows the values 
of ax 2 + bx + c which correspond to the different values 
of x. (4) By inspection determine the points of the curve 
where the expression is zero. The values of x for these 
points are the solutions of the equation. 

EXERCISES 

Solve the following equations graphically, and check : 
1. ic 2 - 9z-f 14 = 0. 

Plot the function between the 
limits to 12. This means con- 
struct the table by letting x equal 
the following values : 0, 1, 2, 3, 
4 ... 12. 



6. 4a 2 

Plot from 2 to 5. 

Plot from - 4 to + 2. 

8. 100 x 2 - 5 x -495 = 0. 
Plot from 5 to 4- 5. 

9. 6a; 2 -17x = 20. 
Subtract 20 from both mem- 



2. z 2 -6z + 5 = 0. 
Plot from - 1 to 7. 

3. x 2 - 3x -10 = 0. 
Plot from - 3 to 6. 

4. x 2 -11 x + 24 = 0. 
Plot from 1 to 10. 

5. x*-llx + 25 = 0. 
Plot from 1 to 10. 

11. 9x 2 + 3x + 20 

Subtract 2 x 2 + 2 x + 50 from both members of the equation. 

529. The graph solves a family of equations. At this 
point the student should note that a single graph may be 
used to solve a whole family of equations. Thus, if we turn 
to Fig. 307 we see that the curve for x 2 2x 360 can 



bers and plot 6 x 2 -11 x- 20. 

10. 2x 2 - 9 = 3x. 

Subtract 3 x from both mem- 
bers and plot the function 
2:r 2 -9 -3z. 



QUADBATIC EQUATIONS 467 

be used not only to solve the equation x* + 2 x 360 = but 
also to solve every equation of the type a? + 2 x 360 = c 
(where c is some arithmetical number). For example, if we 
ask what value of x will make x z + 2 x 360 equal to 100, 
we can tell by looking at the curve that the answer is 20.5 
or 22.5, and this is precisely the same as saying that the 
two roots of the equation j? + 2 x 360 = 100 are 20.5 
and - 22.5. 

EXERCISES 

Solve by the graph : 

1. x 2 + 2x- 360 = 200. 6. x 2 + 2 x = 400. 

2. ** + 2* - 360 =180. 



Subtract 36Q 

3. x- + 2 x- 360 = 400. 

4. & + 2x - 360 = - 250. ? x 2 + 2x - 500 = 0. 

5. x 2 + 2x- 360 = -360. Add 140. Why? 

The last two exercises show that the graph solves all 
equations in which x* + 2 x= c (some arithmetical number). 
For we can write the given equation x 2 + 2 x 500 = in 
the form x* + 2 x 360 = 140. This last form we are able 
to solve at sight by the graph. 

530. The parabola. The curve representing the func- 
tion x*+ 2x 360 shown in Fig. 307 is called a parabola. 
Study and discuss the general shape and symmetry of the 
curve. Compare the curves you and your classmates have 
drawn in the exercises of this chapter and see if you can 
find a parabola in an earlier chapter of this text. 

The graph of a quadratic function in one unknown is 
a parabola. iWis a symmetrical curve. No three points of 
the curve lie on a straight line. The parabola is a common 
notion in physics and mechanics. Thus, the path of a 
projectile (for example, a bullet) is a parabola. A knowl- 
edge of the theory and application of many such curves 



468 GENERAL MATHEMATICS 

was of extreme importance in the recent world war. The 
soldiers who had been trained in some of the more advanced 
mathematical courses, especially in trigonometry and graphi- 
cal work, were in demand and were given plenty of oppor- 
tunity to put into practice what they had learned in school. 

In plotting functions like x? + 2 x 360 we plot the 
values of x along the a>axis and the corresponding values 
of the function on the ?/-axis. This suggests that the 
curve obtained in Fig. 307 is the graph of the equation 
y = z 2 + 2 a: - 360. 

It follows that whenever y is a quadratic function of x, 
or when x is a, quadratic function of y, the graph of the 
equation is a parabola. 

EXERCISES 

Graph each of the following equations : 

1. y = a; 2 -4. 3. z = y 2 + 5 ?/ + 4. 

2. y = x 2 + 3 x + 2. 4. x = f - 1 y + 6. 

531. Maxima and minima. The theory of maxima 
(greatest values) and minima (least values) of functions 
has many important applications in geometry, physics, and 
mechanics. 

This article will present one example drawn from each 
subject. A careful study of the following example will 
suggest the proper method of attack. 

ILLUSTRATIVE EXAMPLE 

A rectangular garden is to be inclosed on ^iree sides, the 
fourth side being bounded by a high wall. WlJft is the largest 
garden that can be inclosed with 20 rd. of fencing ? 

Solution. Let x represent the width. 

Then 20 2 x represents the length, 

and 20 x 2 x 2 represents the area. 



469 



We are now interested in a maximum (greatest possible) value 
that 20 x 2 x 2 can have. By trial we obtain the corresponding 
values for x and the function 20 x 2 x 2 shown 
in the table below. 

Common sense, the table, and the curve of 
Fig. 308 show us that if the garden is made 
very wide or very narrow the area is very small. 
The table and the curve of Fig. 308 suggest that 
50 is probably the largest area. In this case the 
dimensions of the garden are 5 and 10. By 
taking x first a little larger and then a little 
smaller than 5, we may check our conclusion. 



We can as a matter of fact, save our- 



X 


20z 2x 2 








1 


18 


2 


32 . 


3 


42 


4 


48 


5 


50 


6 


48 


7 


42 


10 






selves much of the labor of these computations by an alge- 
braic method, which we shall present in Art. 538. At this 



-/(*> 



-20 



FIG. 308. SHOWING THE MAXIMUM VALUE or A QUADRATIC FUNCTION 

stage, however, we shall be content to plot the curves and 
find the highest or the lowest point on the curve. 



470 GENERAL MATHEMATICS 

EXERCISES 

1. If a ball is thrown upward with a velocity v , the dis- 
tance d from the earth to the ball after a given time t is given 
by the physics formula 

d = v t- 16 1 2 . 

How high will a ball rise which is thrown with an initial 
velocity of 100 ft. per second ? 

HINT. The formula becomes d = 100 1 16 < 2 . Plot the function 
and find by inspection its greatest value. 

2. Divide 10 into two parts such that their squares shall be 
a minimum. 

*3. Find the most advantageous length of a lever for lifting 
a weight of 100 Ib. if the distance of the weight from the 
fulcrum is 2 ft. and the lever weighs 4 Ib. to the foot. 

532. Limitations of the graphic method of solving quad- 
ratic equations. By this time the student is no doubt con- 
vinced that the graphic method of solving quadratic 
equations has its limitations. We may enumerate the 
following: (1) The results are frequently rough approx- 
imations. This is evident the moment we attack problems 
of some slight difficulty. In fact the earlier problems of the 
chapter are artificially built so that in all probability the 
student will accidentally get an accurate result. We must 
remember that the graphic method depends for its accuracy 
upon the mechanical (or nonintellectual) conditions, such 
as the skill of the student at this type of work, the exact- 
ness of squared paper, and our ability to estimate fractional 
parts of the unit. (2) Aside from the fact that the signifi- 
cance of a graph is sometimes obscure, the work is a bit 
cumbersome and tedious. (3) It is not economical of time, 
as we shall presently show. 



QUADRATIC EQUATIONS. 471 

533. More powerful methods of solving quadratic equa- 
tions. Because of the foregoing limitations of the graphic 
method we are ready to proceed to the study of more 
efficient methods. These methods rest purely on an intel- 
lectual basis (that is, the accuracy is independent of 
constructed figures). We shall observe that they get the 
results quickly and with absolute accuracy. 

534. Quadratic equations solved by factoring. The factor- 
ing method may be illustrated by the following solution 
of the speed problem with which we opened the chapter : 

Solution. Given x 2 + 2 x - 360 = 0. 
Factoring the left member, 

(x + 20) (x - 18) = 0. 

The preceding condition, namely, (a; + 20) (z 18) = 0, will be 
satisfied either if x + 20 = or if x 18 = 0, for we learned in 
Art. 236 that the product of two numbers is zero if either factor 
is zero. Thus, 5 x = or x 8 = O/ 

Now if x + 20 = 0, 

then x - 20. 

And if x - 18 = 0, 

then x - 18. 

Hence x - + 20, 

or x-- 18. 

In the next solution we shall omit a considerable part of the dis- 
cussion and show how the work may be arranged in a few simple 
statements. 

Solve z 2 - 9 x + 14 = 0. 

Factoring, (x - 7) (x - 2) = 0. (1) 

This equation is satisfied if x 7 = 0, (2) 

or if x - 2 = 0. (3) 

From equation (2) x = 7, 

and from equation (3) x = 2. 

The numbers satisfy the equation, consequently 2 and 7 are the 
roots of the equation x' 2 9 x + 14 = 0. 



472 GENERAL MATHEMATICS 

EXERCISES 

Solve the following quadratic equations by the method of fac- 
toring and test the results by substituting in the equations : 

1. x 2 - 5x + 6 = 0. 15. 20x - 51 = x 2 . 

2- if -7 y +12 = 0. 16. 77+4d = d 2 . 

4. x 2 - 6x-27=0. 18.10^ + 33^=7. 

. 5. x 2 -2x-35 = 0. 19. 6z 2 =23z + 4. 

6. cc 2 + 5x = 6. x I I 

20. + = 
HINT. Subtract 6 from '2 2 x 

both members before apply- 3 

ing the method. Why? 21> 15a; + 4 = ^' 

7. ar + a; = 56. r _ jt 33 

o-l 

HINT. Subtract 2x from oo K ~ Q _. 

&O> t-J *)U O ^~~ 

both members and rearrange x 

terms before factoring. y 15 ?/ 2 

9. ^ + 4 = 42,. " 2 + T" = 14' 

10. x 2 - 85 =12 a-. 25.-^-+^ =4. 

11. 2 =10^ + 24. ~ y 

12. ?/? 2 91 = 6 w. 26 - 2 = "3 a + 3' 

XO* i-C ~~ r~ 3C - -r*-. /y or 

07 x _ i _ o 

14. m 2 +112 = 23m. ' x-3 a; + 3 

VERBAL PROBLEMS 

Solve the following problems by the factoring method and 
test the results by substituting the solution in the conditions 
of the problem : 

1. A crew rows across a calm lake (12 mi. long). On the 
return trip it decreases the rate by 1 mi. per hour and makes the 
trip in 7 hr. Find the rate of the crew both going and returning. 



QUADRATIC EQUATIONS 



473 



2. A man drives 'a car 80 mi. out of town. On the return 
trip he increases his rate 8 mi. per hour. He makes the trip 
in 4^ hr. What was his rate while driving out ? 

3. The base of a rectangle exceeds the altitude by 5 in. The 
area equals 150 sq. in. 



Find the 
altitude. 



base and 



40 



-90-2W- 



T 

I 

40 -2W 

f 

W 

I 



90 
FIG. 309 



4. The base of a 
triangle is 7 in. less 
than the altitude. The 
area equals 85 sq. in. 
Find the base and 
altitude. 

5. A farmer is plowing a field of corn 40 rd. wide and 90 rd. 
long (Fig. 309). At the end of a certain day he knows that he 
has plowed five sixths of the field. How 

wide a strip has he plowed around the 
field? 

6. A piece of tin in the form of a 
square is taken to make an open box. 
The box is made by cutting a 1-inch 
square from each corner of the piece of 
tin and folding up the sides (Fig. 310). 
The box thus made contains 36 cu. in. 

Find the length of the side of the original piece of tin. 

535. Limitations of the factoring method. In some of the 
preceding exercises the quadratic expressions were very 
difficult to factor. This is usually the case when the 
constant terms are large numbers. Indeed, most verbal 
problems lead to quadratic equations which cannot be 
solved by the factoring method. The following problem is 
a simple illustration: 



FIG. 310 




474 GENERAL MATHEMATICS 

What is the length of a side of a square (Fig. 311) whose 
diagonal is 2 ft. longer than a side ? 
Attempted solution by factoring method: 
Let x = a side of the square. 

Then x + 2 = the length of the diagonal, x 
By the theorem of Pythagoras 

x z + x 2 = x 2 + 4 x + 4. 
Simplifying, & - 4 x - 4 = 0. 

This appears to be the end of the road ; 

we cannot factor x 2 4 x 4, for we cannot obtain a combination 
of whole numbers or fractions whose product is 1 and whose sum 
is 4. And yet we are probably convinced that such a square does 
exist though we are forced to admit that the solution of the problem 
by the factoring method is hopeless. 

536. Solution of the quadratic equation by the method of 
completing the square. If we were able to make the left 
member of the equation z 2 4 a; 4 = a perfect square 
without introducing the unknown (#) into the right mem- 
ber, we could take the square root of each member of the 
equation and thus obtain a linear equation which would 
be easily solved. This is precisely the method we wish to 
employ. However, we must first learn to make the left 
member x 2 - 4 x 4 a perfect square. 

ORAL EXERCISES 

1. Find (x + 2) 2 ; (x + 3) 2 ; (x + 4) 2 ; (a; - 2) a . 

2. When is a trinomial a perfect square ? (See Art. 250.) 

3. Make a perfect square trinomial of the following: a- 2 6 x; 



7 x 
x 2 + 7 x ; x 2 + 9 x ; x 2 + 



QUADRATIC EQUATIONS 475 

The preceding exercises show that it is easy to complete 
the square of a binomial of the form x 2 + ax, for we need 
only to add the square of half the coefficient of x. Then, too, 
the constant term of a trinomial can always be made to 
appear in the right member of the equation, leaving the 
left member in the form x 2 + ax. We now proceed , to 
solve the equation x 2 4 # 4 = 0. Write the equation 

/ 4\ 2 
thus: X s 4 # = 4. Add (5-) or 4, to make the first 

\ 2 / 

side a trinomial square, and we obtain 
x z _ 4 x _|_ 4 = 8. 

Taking the square root of both sides and remembering 
that 8 has two square roots, + V8 or V8, we get 



(1) 
or ar-2=-V8. (2) 

From equation (1) we get x = 2 + V8, and from equation 
(2) we get a:=2_-V8. 

If we obtain V8 either by the arithmetical method taught 
by Art. 446 or by using the table of Art. 449, the result 
(accurate to three places) is 2.828. Then x= 2 + 2.828, or 
4.828. Hence the side of the square whose diagonal is 2 ft. 
longer than a side is 4.828 ft. We can check this result by 
applying the theorem of Pythagoras. 

We reject 2 V8, or 0.828, because it does not satisfy 
the conditions of the problem. However, the student should 
realize that 0.828 is just as much a solution of the 
equation z 2 4 x + 4 = 8 as is 4.828. 

The method of completing the square is further illus- 
trated by the following solution of the equation 

10 z 2 - 9* + 2 = 0. 



476 GENERAL MATHEMATICS 

Write the equation 10 x 2 - 9 x = - 2. Why '? 

9 x 1 

Dividing by 10, z 2 - = - - . 

1U 5 

Note that, the left member is now more easily completed. Why ? 

x _ 

10 

x2 _9 + ^L-.J_. Why? 

10 + 400 400 

Taking the square root of each member, 
x - & = 5 V 
Whence x = % or f . 

537. Summary of the method for solving quadratic equa- 
tions by the method of completing the square. 

1. /Simplify the equation and reduce to the form ax* + 
bx = c. 

2. If the coefficient of x 2 is not 1, divide both members of 
the equation by the coefficient so that the equation takes the 
form x 2 +px = q. 

3. Find half the coefficient of x; square the result ; add 
the square to both members of the equation obtained in step 2. 
This makes the left member a perfect square. 

4. Express the right member in its simplest form. 

5. Take the square root of both members, writing the double 
sign before the square root in the right member. 

6. Set the left square root equal to the positive root in the 
right member of the equation in 5. Solve for the unknown. 
This gives one root. 

7. Repeat the process, using the negative root in 5. This 
gives the second root of the equation. 

8. Express the roots first in simplest form. 



QUADRATIC EQUATIONS 477 

EXERCISES 

Solve by the method of completing the square, and check : 

' 1. x a -6 = 91. 11. 4z 2 + 45z-36 = 0. 

2. x 2 -8a; = 48. 12. 6z 2 + 7cc-20 = 0. 

3. X 2_ x _3 = o. 13. 2 2 + 62=l. 

4. y 3 _|_ 4y _)_ 3 = 0. HINT. Compute roots to the 

- o i A K n nearest hundredth. 

5. if + 4y 5 = 0. 

6. /,* _(- 8 b - 20 = 0. 14. x 2 + 4 a = 16. 

7. y 2 +14^-51 = 0. 15- z 2 = 24 + 4a:. 
8. m 2 + 5 wi - 6 = 0. 16. 7 + 2x = a; 2 . 

9. a; 2 -13z + 40 = 0. 17. 75-3x 2 =75. 

10. x 2 + 6x + 5 = 0. 18. 19 - a = 4 a 2 . 

VERBAL PROBLEMS 

1. If 4 is taken from a certain number the result equals 96 
divided by that number. Find the number. 

2. Find the two consecutive numbers the sum of whose 
squares equals 113. 

3. In physics we learn that the distance in feet which a 
stone thrown downward goes in a given time equals 16 
multiplied by the square of the number of seconds it has fallen, 
plus the product of the velocity with which it is thrown and 
the number of seconds fallen ; that is, s = vt + 16 1*. Suppose 
that v = 20 ft. per second and s = 1800 ft. Find the value 
of t. Try to state the meaning of this problem in simple 
(nontechnical) words. 

4. How long will it take a baseball to fall from the top of 
the Washington Monument (555 ft.) if it starts with a velocity 
of 50 ft. per second ? 

HINT. Solve the equation 16 t 2 + 50 t = 555. 



478 GENERAL MATHEMATICS 

5. How long will it take a body to fall 800 ft. if it starts at 
20 ft. per second ? 

6. How long will it take a bomb to fall from a Zeppelin 
1000 ft. high if it starts with no initial velocity ? 

7. Two trains are 175 mi. apart on perpendicular roads and 
are approaching a crossing. One train runs 5 mi. an hour faster 
than the other. At what rates must they run if they both 
reach the crossing in 5 hr. ? 

NOTE. The 175 means the distance along the track. 

8. The circumference of the fore wheel of a carriage is less 
by 3 ft. than the circumference of the hind wheel. In traveling 
1800 ft. the fore wheel makes 30 revolutions more than the 
hind wheel. Find the circumference of each 

wheel. 

9. A window (Fig. 312) in the form of a 
rectangle surmounted by a semicircle is found 
to admit the most light when the width and 
height are equal. If the area of such a window p 
is 175 sq. ft., what is its width ? 

10. A boy has a piece of board 16 in. square. How wide a 
strip must he cut from each two adjacent sides to leave a 
square piece whose area is three fourths that of the original 
piece ? In what form would you state your result to meet all 
practical purposes ? 

11. A lawn is 30 ft. by 80 ft. Two boys agree to mow 
it. The first boy is to mow one half of it by cutting a 
strip of uniform width around it. How wide a strip must 
he cut? 

12. A farmer has a field of wheat 60 rd. wide and 100 rd. 
long. How wide a strip must he cut around the field in order 
to have one fifth of the wheat cut ? 




QUADKATiC EQUATIONS 479 

13. In a circle of radius 10 in. the shortest distance from 
a given point on the circumference to a given diameter is 8 in. 
Find the segments into which the 

perpendicular from the point divides 
the diameter. 

HINT. Study Fig. 313 and try to re- 
call the various theorems we have proved 
involving mean proportional. If you fail 
to get a solution, refer to the mean pro- 
portional construction (Art. 374). 

14. A broker sells a number of- rail- 
way shares for $600. A few days 

later, the price having risen f 10 a share, lie buys for the same 
sum three less shares than he sold. Find the number of 
shares transferred on each day and the price paid. 

15. A boy sold a bicycle for $24 and lost as many per cent 
as the bicycle had cost him in dollars. Find the cost. 

16. A line 20 in. long is divided into two parts, A C and CB, 
so that AC is the mean proportional between CB and AB. 
Find the length of A C. 

*538. Maxima and minima algebraically determined. We 
have seen in Art. 531 that the graph of a quadratic func- 
tion may be used to determine the maximum or minimum 
values of the function. This method is not as exact, how- 
ever, as the algebraic method of determining maxima and 
minima, and it is longer. Take, for example, the problem 
of Art. 531 to find the maximum value for 20 x 2 a- 2 . 
If we represent this maximum value by m, then we may 
write 20 x 2 a- 2 = m. If we divide both sides of the 
equation by 2 and complete the square on the left side 
of the equation, 

(x - 5)2= 25 -|. 



480 GENEKAL MATHEMATICS 

It is evident from this equation that if x is a real 
number, m cannot be greater than 50. Therefore the maxi- 
mum value of m, or of 20 x 2x*, is 50. 

In like manner by letting m represent the minimum 
value of a function we can determine when in is a mini- 
mum more quickly and more accurately than we can by 

the graphic method. 

EXERCISES 

Determine the maximum or minimum values of the follow- 
ing functions : 

1. 3x a -4ar-l. 3. ->x*-x-l. 5. x* - 6x + 8. 

2. 2 + 2Z-2X 2 . 4. 1-x 2 . 6. 6-x-x 2 . 



SUMMARY 

539. This chapter has taught the meaning of the follow- 
ing words: parabola, maxima and minima. 

540. This chapter has taught three methods of solving 
a quadratic equation of one unknown: the graphical 
method, the factoring method, and the method of com- 
pleting the square. 

541. The graphical method proved to be the most con- 
crete. It presented a clear picture of the changes hi the value 
of the function which correspond to changes in the value of 
the unknown. It also served as a sort of " ready reckoner." 

542. The factoring method was more expedient; the 
results were obtained by this method much more quickly 
and with greater accuracy. 

543. The method of completing the square was used 
to solve quadratic equations which were not solvable by 
the method of factoring. 

544. Both the graphical and algebraic methods of de- 
termining maxima and minima were presented. 



INDEX 



Absolute value, 156, 161 
Acute angle, 49 
Addend, 39 
Addition, algebraic, 162 

commutative law, 38 

geometric, 36 

law of, 4 

of monomials, 160-163 

order of terms in, 38 

of polynomials, 168 

of positive and negative numbers, 

160-169 

Adjacent angles, 60 
Ahmes, 88 
Ahmes Papyrus, 360 
Algebra, origin of word, 8 

"shorthand" of, 13 
Alloy problems, 338 
Altitude, 67 
Angle, acute, 49 

bisection of, 68 

central, 52 

complement of, 119 

construction of, 56, 62 

definition, 47 

degree of, 55 

of depression, 352 

of elevation, 350 

initial side of, 48 

left side of, 114 

measurement of, 54, 66 

negative, 159 

notation for, 50 

obtuse, 49 

positive, 159 



Angle, reflex, 49 

right, 48 

right side of, 114 

straight, 48, 111, 112 

supplement of, 116, 118 

symbols for, 48 

terminal side of, 48 

vertex of, 47 
Angles, adjacent, 60 

alternate exterior, 127 

alternate interior, 124 

comparison of, 59 

complementary, 119 

corresponding, 68 

exterior, 139 

geometric addition of, 61 

geometric subtraction of, 61 

interior, 125, 130 

positive and negative, 159 

size of, 48 

supplementary, 116, 118 

supplementary adjacent, 111, 
116 

vertical, 122 
Arc, degree of, 65 

intercepted, 52 
Area, measurement of, 74 

of a parallelogram, 79 

practical method of estimating, 74 

of a rectangle, 75 

of a rhombus, 81 

of a square, 78 

of a trapezoid, 82 

of a triangle, 81 

unit of, 74 



481 



482 



(JKNKKAL .MATIIK.MATK'S 



Areas, calculating, 205 

proportionality of, 341 
Arithmetic average, 244, ii4o, 247 
Arrangement, 167 
Ascending powers, 167 
Axioms, 21, 22, 37 

Bar diagram, construction of, 224 

interpreting, 222 
Base, 102 

Beam problems, 336, 385 
Bearing, of a line, 353 

of a point, 354 
Bhaskara, 108 
Binomial, cube of, 416 

geometric square of, 91 

square of, 92 
Bisector, construction of, 66, 68 

perpendicular, 66 
Braces, 175 
Brackets, 175 
Briggs, Henry, 59, 447 

Cartograms, 230 
Centigrade, 289 
Central tendencies, measures of, 

244 
Characteristic, 428 

table of, 437 
Circle, arc of, 52 

center of, 52 

circumference of, 52 

construction of, 51 

definition of, 51 

degree of arc of, 53 

diameter of, 52 

quadrant, 52 

radius of, 52 

semicircle, 52 
Class interval, 239 
Class limits, 240 



Coefficient, 39 
Coincide, 34 

Commutative law, 38, 85 
Compasses, 31 

measuring segment, 32, 33 
Compensating errors, 255 
Complement of an angle, 119 
Consecutive-number problems, 13 
Constant, 301 
Coordinates, 265, 368 
Corresponding parts, 346 
Cosine, 357, 358, 361 
Cube, 98 

Cube root, of arithmetical num- 
bers, 420 

by slide rule, 456 

by table, 397, 398 
Cumulative errors, 255 
Curve, normal distribution, 257 

skewness of, 260 

symmetry of, 259 

Data, 214 

Decagon, 44 

Decimal point, logarithms, 435 

Degree, of angle, 53 

of arc, 53 

of latitude, 53 

of longitude, 53 

of a number, 166 
Dependence, 300 
Dependent variable, 300 
Depression, angle of, 352 
Descartes, 108 
Descending powers, 167 
Difference, of monomials, 42 

of two-line segments, 38 
Direct variation, 308 
Dissimilar terms, 7 
Distance, 280 
Division, checking long, 209 



INDEX 



483 



Division, definition of, 194 
law of signs in, 195 
of monomial by monomial, 105 
of negative numbers, 194 
of polynomial by monomial, 197 
of polynomial by polynomial, 207 
with slide rule, 454 
by zero, 211 

Drawing to scale, 345-355 

Elements of geometry, 88 
Elevation, angle of, 350 
Elimination, 373 

by addition or subtraction, 374, 
375 

by substitution, 377 

summary of methods of, 379 
Equal segments, 34 
Equation, checking, 6 

definition and properties, 12 

laws for solving, 2-5, 9 

members of, 2 

number satisfies, 6 

quadratic, 462 

root of, 7 

substituting in, 6 

translation of, 12 
Equations, contradictory, 371 

dependent, 372 

equivalent, 372 

exponential, 443, 448 

identical, 372 

inconsistent, 372 

indeterminate, 370 

outline of systems, 373 

pupils' test of simple, 243 

simultaneous linear, 367, 369 

system of, 369 

systems containing fractions, 381 

in two unknowns, 373 
Equiangular, 149 



Equilateral, 44 

Euclid, 88 

Evaluation of formulas, 93, 279, 

290 
Exponent, indicates degree, 166 

logarithm as an, 427 
Exponents, 102, 425 

fractional, 412 

negative, 415 

zero, 414 
Extremes, 323, 328 

Factors, 198 

common monomial, 198 

"cut and try," 199 

of difference of two squares, 202 

equal, 102 

prime, 200 

of trinomial square, 202 
Fahrenheit, 289 
Fallacious proof, 211 
Formula, definition, 78, 273 

evaluating, 93, 279, 290 

interest, 273 

motion problem, 279 

solving, 276, 294 

summary of discussion, 279 

translating into, 287 

work problem, 283-285 
Fourth proportional, 334 

construction of, 334 
Frequency table, 239 
Fulcrum, 336 
Function, 359 

defined, 299 

dependence, 300 

graph of, 301 

graphical solution, 304 

linear, 303 

quadratic, 463 
Functions, trigonometric, 359 



484 



GENERAL MATHEMATICS 



Geometric problems, 376 
Geometry, origin of word, 88 
Graph, 215 ' 

of centigrade, 290 

of constant-cost formula, 261 

of data, 214 

of Fahrenheit, 290 

of functions, 301 

of linear equations, 263-269 

of quadratic equations, 463-469 

of simultaneous linear equations, 
369 

of variation, 308-311 

of x = 10*, 442 

of y = x 2 , 391 
Graphic curve, construction of, 233 

interpreting, 231 
Graphing, terms used in, 267 

Hexagon, 44 
Hipparchus, 360 
Hyperbola, 311 

Identity, 204 

Independent variable, 300 

Index, 420 

Indirect measurement, 345-355 

Inequality, 34 

Inertia of large numbers, 255 

Intercepted arc, 52 

Interest formula, 273-275 

involving amount, 277 
Interest problem, graphical solu- 
tion of, 276 

solved by logarithms, 444 
Interpolation, 437 
Intersection, point of, 26 
, Inverse variation, 309-311 
Is8sceles triangle, 138 

Joint variation, 312 



Labor-saving devices, 424 

Latitude, 63 

Law of the lever, 336 

Laws, 21, 38, 85, 179, 196, 336 

Least common denominator, 324 

Least common multiple, 11 

Length, 26 

measurement of, 27 

units of, 28 
Lever arm, 181 
Line, bearing of, 353 

of sight, 351 
Line segments, 26 

difference of two, 38 

equal and unequal, 34 

ratio of, 35 

sum of, 36 
Linear equations, definition, 266 

graphic solution of simultaneous, 

369 

Locus, 266 
Logarithm, of a power, 433 

of a product, 432 

of a quotient, 434 
Logarithms, 424, 450, 451 

applied to slide rule, 449 

historical note on, 446 

notation for, 427 

position of decimal point, 435 
Longitude, 53 

Mannheim slide rule, 450, 453 
Mantissas, 428 

table of, 430-431 
Maxima, 468-469, 480 
Mean proportional, 329 
Means, 323, 328 
Measurement, of angles, 54 

of areas, 74 

errors in precise, 30 

indirect, 345-355 



INDEX 



485 



Measurement, of line segments, 27 

of volumes, 99 
Median, 66, 250 

construction of, 67 

how to determine, 251 
Members of an equation, 2 
Metric system, 28 

advantages of, 29 

historical note on, 29 
Minima, 468-469, 480 
Minutes of angle, 53 
Mixture problems, 338, 379 
Mode, 248 

advantages of, 249 

disadvantages of, 249, 250 
Monomial, degree of, 167 

factors of, 198 
Monomials, division of, 195 

multiplication of, 184 

sum of, 177 
Motion, circular, 284 
Motion problems, 279, 384 

graphical illustration of, 283 
Multiplication, abbreviated, 95 

algebraic, 89-90 

by balanced bar, 180 

commutative law of, 85 

law of signs in, 179 

of monomials, 184 

of a polynomial by a monomial, 
186 

of positive and negative num- 
bers, 178-194 

several factors, 184 

with slide rule, 453 

special products, 192, 193 

of two binomials, 190 

of two polynomials, 187, 188 

by zero, 183 

Napier, John, 446 



Negative number, 151 
Newton, Isaac, 418 
Normal distribution, 257, 268 
Notation, for angles, 50 

for logarithms, 427 

for triangles, 130 
Number, absolute value of, 156 

algebraic, 151 

degree of, 167 

literal, 6 

negative, 150-157, 178-180 

numerical value of, 156 

positive, 150-154 

prime, 198 
Number scale, 170 
Number-relation problems, 378 
Numbers, difference of algebraic, 
177 

ratio of, 35 

representation of, 152-153 

sum of algebraic, 177 

Obtuse angle, 49 
Octagon, 44 
Opposite angles, 122 

of parallelogram, 142 
Order of powers, 167 

Parabola, 467 

Parallel lines, construction of, 69 

definition, 68 
Parallelepiped, 99 

volume of oblique, 100 

volume of rectangular, 99 
Parallelogram, defined, 70 

opposite angles, 142 
Parenthesis, 175-176 
Partial products, 89 
Pentagon, 44 

Perfect trinomial square, 393 
Perigon, 48 



480 



GENERAL MATHEMATICS 



Perimeter, 43 
Perpendicular, 65 

bisector, construction of, 66 
Pictograms, 214-220 
Point, bearing of a, 354 

determined by, 26 
Polygons, classified, 44 

equilateral, 44 

similar, 318 
Polynomial, 40 

degree of, 167 
Polynomials, addition of, 168 

classified, 40 

division of, 208 

multiplication of, 187 

subtraction of, 173 
Positive numbers, 151-154 
Power, 103 
Powers, ascending, 167 

descending, 167 

table of, 397, 398 

of ten, 425-426 
Prime number, 198 
Problems, alloy, 338 

beam, 337, 385 

clock, 284 

consecutive-number. 13 

digit, 378-379 

geometric, 376 

interest, 273 

lever, 336 

mixture, 338, 379 

motion, 279, 384 

number-relation, 378 

recreation, 386-388 , 

specific-gravity, 340 

work, 285 
Product, accuracy of, 93 

geometric, 89 

law of order, 85 

monomial, 85 



Product, partial, 89 
of a polynomial and a monomial, 

86 

of powers, 103 
of two polynomials, 187 

Proportion, 322 
beam problems, 337, 385 
different arrangements of, 327 
mean proportional, 329 
means and extremes of, 323, 328 
mixture and alloy problems, 338 
specific-gravity problems, 340 
test of proportionality, 324, 328 

Proportional, constructing a mean, 

332 

fourth, 334 
inversely, 309 

Protractor, 54 

Pyramid, frustum of, 98 
triangular, 98 

Pythagoras, 401 

historical note on, 402 
theorem of, 397, 399, 400 

Quadrant, 52 
Quadratic equation, 463 
Quadratic equations, completing 
the square, 475-477 

factoring method, 471-474 

graphic method, 465 

two solutions of, 465 
Quadratic function, 463 

graph of, 463 

Quadratic surd, 392, 406-407 
Quadratic trinomial square, 392 
Quadrilateral, 44 
Quotients in per cents, 335 

Radical expression, 420 
Radical sign, 390 
large number under, 406 



INDEX 



487 



Radicand, 390 

Rate, 280 

Ratio, 35, 315, 345 

trigonometric, 359 
Rationalizing denominator, 409 
Rectangle, 70 
Reflex angle, 49 
Regiomontanus, 360 
Removal of parenthesis, 175-177 
Rhombus, 81 
Right angle, 48 
Right triangle, 135 
Right triangles, similar, 355 

sum of acute angles in, 137 
Root of an equation, 6 
Roots, fractional exponents, 412 

higher, 420 

by logarithms, 439 

table of, 397-398 

Scale drawings, 345-355 
Seconds, of an angle, 53 
Sector, 217 
Segment, line, 26 

unit, 27 
Semicircle, 52 

Series, continuous and discrete, 237 
Signs, 151 

change of, 171 

law of, 179, 195 

minus, 151-152 

plus, 152 

Similar terms, 40, 42 
Similar triangles, 315 

construction of, 314-316 

corresponding sides of, 315, 330 
Similarity, 345 

Simultaneous equations, 367, 369 
Sine of an angle, 357 
Slide rule, 449 

decimal point, 458 



Slide rule, index of, 452 

inverted slide, 457 

proportion problems, 457 

raising to powers, 455 

runner, 454 

Solids, geometric, 98, 99 
Solving equations, 6 
Specific gravity, 340 
Sphere, 98 
Square, 71 

of the difference, 192 

of the sum, 192, 193 

trinomial, 202 
Square root, 390-397, 451 

algebraic rule for finding, 396 

of a fraction, 408, 409 

by graphic method, 391 

by logarithms, 439-440 

by mean proportional, 405 

memorizing, 408 

of a product, 406 

by ruler and compass, 404 

by slide rule, 455 

table of, 398 
Squared paper, 32 
Statistical regularity, law of, 

254 
Statistics, 214 

defined, 238 

historical note on, 270, 271 

limitations of, 253 

use of, 238 
Steel tape, 349 
Straight angle, 48 
' Substituting, 6 
Subtraction, algebraic, 171 

graphical, 38, 170 
Sum, of angles about a point, 
112-113 

of angles of a triangle, 131 
Supplement, 116, 118 



488 



GENERAL MATHEMATICS 



Surds, addition and subtraction of, 

410 

multiplication and division of , 41 1 
Surveying, 58 
Surveyor's chain, 349 
Symmetry, 259 
Systems of equations, 369 

Table, of mantissas, 430-431 

of roots and powers, 397-398 

of trigonometric ratios, 361 
Tangent of an angle, 358 
Tape, 349 
Terms, 7 

dissimilar, 7 

order of, 7 

similar, 7 

Test of proportionality, 324 
Tetrahedron, 106 
Theorem of Pythagoras, 397-399 
Transit, 58 
Transversal, 68 
Trapezoid, 82 
Triangle, altitude of, 67 

area of, 81 

base angles of, 138 

base angles of isosceles, 138 

defined, 43 

equilateral, 44 

exterior angles of, 139 

isosceles, 138 

notation for, 130 

perimeter of, 43 

right, 135 

sides of, 43 

similar right, 317 

solving, 359 

sum of exterior angles, 139 

sum of interior angles, 131-135 

trigonometric area formula, 365 



Triangle, vertex, 47 

vertices, 43 

wooden, 136 
Triangles, construction of, 142-147 

similar, 314 

Trigonometric ratios, 359 
Trigonometry, 356 
Trinomial, 40 

factoring, 199 
Trinomial square, 202 
Turnin : tendency, 181 

Unequal angles, 60 

Unequal lines, 34 

Units of measure, 28, 74, 99 

Value, absolute, 156 
Variable, 300 

dependent, 300 

independent, 300 
Variation, direct, 305 

inverse, 309 

joint, 312 

Vel Deity or rate, 280 
Verbal problem, method of solving, 

16 

Vertex of an angle, 47 
Vertical angles, 122 
Vieta, 108 
Vinculum, 175 
Volume, of cube, 102 

measurement of, 99 

of parallelepiped, 100 

unit of, 99 

Work problems, 285, 383 

Zero, division by, 211 
multiplication by, 183 



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